The Whole of A Level Maths | Pure | Revision for AQA, Edexcel, OCR AND WJEC

hey guys i am really excited about this because this is the start of everything you need to know for a level math this is the pure section and the lovely mystery is going to be taking you through each skill one at a time for every single skill in here there is a much longer video going through each skill at three different levels lots and lots of questions so if there is one bit that you're not too sure about go and watch the longer video is explained and lots of examples are worked through to help you ensure that you can do each of these skills over my website there are thousands of multiple choice questions just waiting to help you revise and make sure that you've got the confidence in these skills so you can go into the exam and apply it when needed [Music] the laws of indices normally involve a base number let's call that n and something happening to the base number so for example then being multiplied together now with the laws of indices we also have an index involved so we might have the power a on our first n and the power b on our second and so the law is that if you're multiplying together the base number and to simplify it you can add the powers together so let's have a look of an example of this in action so you might have 7 to the power of 2 multiplied by 7 to the power of 3 and that can be simplified to 7 to the power of 5. now there's laws like this for subtraction as well and that's when you're dividing the base numbers so you're dividing the base numbers then the two indexes the indices they are going to be taken away so again looking at an example you could have 3 to the power of 5 divided by 3 to the power of 2 and that will simplify to 3 to the power of 3 taking the powers away we also have a version for multiplication we have your base number and you multiply a and b and what that looks like is you actually have one base number in this situation it's going to lead to a power a and then all of that is going to be to a power b so if you've got a bracket and you've got basically two powers you're gonna multiply the powers together so looking at another example let's say we've got 4 to the power of 2 and then that is all to the power of 4 then all together that's going to give us 4 to the power of 8. we can also have one where you divide the powers and rather than writing this as a division it's probably going to be um more commonly seen as a fraction instead but fractions and divides are the same thing now let's say that a is the numerator and b is the denominator so how would this look like well we're going to have our base number it's gonna be to the power of a and then it's going to be to the root b so if i look at an example if we have the cube root of a number so let's say this number is going to be a2 and 2 is to the power of let's say 6 then that would be equal to 2 to the power of 6 over 3. and then of course you can simplify that 6 divided by 3 is 2 so that would end up being 2 to the power of 2. so always look for opportunities to simplify these fractions if you can now there are a few extra details of laws of indices so for example if you have a number the power of one then that should stay as writing the number if you have a number to the power of zero then that's equal to 1. if you have a number to a negative power then that's the same as 1 over that number to the positive power so looking at an example of this if we had say 3 to the power of negative two that will be equal to one over three to the power of two and actually three to the power two is uh nine so that'll be equal to one over nine so just like you can simplify the fractions uh you can also simplify fraction of these cases and actually multiply out things if it's going to make it simpler but for now we'll just focus on the core rules now i've used numbers for all my examples but you can use algebra so for example you could have a to the power of 2 multiplied by eighth and that will be a to the power of five or you could have a to the power of two with brackets and then you could have say a b on the outside and so that would be a to the power of and that'd be 2 times b is 2b so we can have algebra involved we can also have a mix of letters and numbers so for example we could have 3a squared times 4a squared so use for the normal rules than all the numbers three times four is twelve then you would follow the laws of indices for the indices two plus two is four so three a squared times four a squared would be twelve a squared three times four is twelve two plus two is four if this happened with the brackets be really careful so you could have for example two a to the power of two all to power three now you don't just do two times three or six and write two a to the power of six that's the wrong answer because not only is the a being raised to the power of three so you do two times three or six the two is being raised to the power of three and two to the power of three is eight so be careful of that you can also have multiple letters so you could have a to the power of two and b to the power of three multiplied by a and then b to the power of four so you do the letters separately a two and a that's two plus one is a square three notice there's no one on the a well as i said before anything to the power of one is just that number then do the same thing with b's three plus four is seven so we can have a combination like that the last thing to watch out for is negative numbers you can have double negatives and those four normal rules double negatives you'll particularly find when you're dividing because we definitely take away the powers and sometimes one of the powers is negative so be really careful double negatives there and the other one is with the fraction when you've got n to the power negative a you can turn that into a fraction or you make it positive this can happen also with a positive m to the power of a would be 1 over n to the power of negative a so you're not really making the fraction positive what you're doing is you're flipping the sign on the power if you're turning it into a fraction one last thing to look at is all of these have the same base number now let's say the base numbers are different so let's say we have seven to the power of two multiplied by two to the power of 3. now because the base numbers are different than seven other two rather than a seven and a seven are a two and a two then in that situation you can actually uh simplify it what you could do though is say that well seven to the power two is forty nine and 2 to the power 3 is 8. which will give you 49 times 8 and that will give you all together 392. so if you have the same base numbers you can do simplification if you don't have the same base numbers the best you can hope for is change them into the values and actually just get a value for it the same thing also goes with addition and subtraction with these we know we've got loss of multiplication division brackets and roots we don't have any rules for adding together the same base number or subtracting the piece number and that's cause we're answering the rules and again like the example just did if you're adding or subtracting the best you can hope for is turning them into the values and working out that way we have a few laws with certs so for example if you have a root let's say root a and you multiply it by let's say root b then you can simplify that into the square root of a times b and the same thing will work with division if you got the root of a divided by the root of b or written as a fraction that's the same thing as the root of a divided by b or again written as a fraction so let's look at some examples so let's say we had the square root of nine multiplied by the square root of seven then we could say that all together that's equal to the square root of 63 if we had the square root of a hundred over the square root of four then we could say that is equal to the square root of a hundred divided by four and let's just write these out with a little bit more detail so then that would be the square root of a hundred divided by 4 which is the square root of 25. you can also have it laws for addition and taken away now these are with the same base root so we have say a lot of root n plus b lots of root n then that would give us a plus b lots of root n now the general term looks pretty confusing but let's have a look at an example so if we had three lots of root seven plus two lots of group seven then all it means is that would give you five lots of root seven and the same thing would work with subtraction now you can also simplify sets so for example the previous answer we got root 25 well we can find the square root of root 25 25 is a square number 5 times 5 is 25. another way we can simplify is in the first question we had root 9 times root 7. there's another way of writing that root 9 we can work it out root 9 is three so we could have written three times root seven which would have given us three root seven so that would be a lot simpler than writing root 63 we're using smaller numbers and in fact what you can do is work backwards you might be given root 63 as a question and you have to recognize that 63 is a square number multiplied by something else it's 9 times 7 and then you can change the 9 into the 3 and then you can write your 3 root 7 that way so that first example you can also be working backwards through that also with sirs we have something called rationalizing the denominator so let's say we had 33 over root 11. that looks okay but we don't like to have roots on the bottom of a fraction so what we do is using equivalent fractions we take the 33 over root 11 and we would multiply the denominator by root 11 and the numerator by root 11. now why we multiply by those numbers in particular well if you multiply a root by itself you're going to get just that number so root 11 times root 11 you're effectively squaring it and you square root it cancels out so that leaves you a number over 11. and then on the numerator we're going to have 33 lots over 11. and so that's a better form to have because with a lot of maths you don't want a irrational number for your denominator so you might be asked to get rid of it and let's just make a note at the top if you've got say root a multiplied by root a then that will give you a another thing to watch it for rationalizing the denominator is that we can actually do the division here we've got 33 over 11. 33 divided by 11 is three so that will give us three lots of root 11. another thing to look out for when rashford denominator you might have something like this and the denominator might be some brackets with a certain it and with a number with it now with this you can't just multiply through by root three because you multiply the one by root three which will leave you the root three at the bottom so again we'll just take our numbers it might even be that you see the question without the brackets actually when you're doing the working eight put the brackets in and you want to multiply it by a negative version of the same bracket now what that does is when you expand this quadratic and you can have quadratics with serves and just be careful when you're multiplying a third by a number you know root three times one is one root 3 and root 3 times root 3 will give you just 3. so be careful things like that but what this would do is for the numerator we'd have 2 times root 3 and 2 times negative 1 so that's fairly simple but with a quadratic you're going to do you've kind of four terms of multiplication root three times root three equals three three times negative one will give negative one root three one times root three will give us a positive root three and then one multiplied by negative 1 will give us negative 1. now when you get this what you'll notice is the middle two terms will cancel out to give you zero and you're also probably going to have two number terms in the denominator so the numerator will stay 2 root 3 minus two well the denominator the negative root three and the positive root three cancellate then we have three taking away one is two and then we can even simplify that because we can actually divide both terms by two two lots of root three divided by two is 1 root 3 and negative 2 divided by 2 is negative 1. now you can't always do that but always look at situations where you can do that when expanding brackets we'll go through a few examples so you might have something like a linear expansion where we have one bracket for example two lots of three x plus one and the way to do this is multiply the two by the three x which will give us six x and then multiply the two by one which will give us 2. it's really important to multiply by both because the common mistake is only multiplying by one of those you might have a letter on the outside like x and on the inside you might have 4x plus the letter y so again the same method x times four of x look at the numbers first of all we've only got four then look at the letters we've got two lots of x that's x squared then we do x times y and we can't do x times y we don't know the values for them so we can't do something you just leave it as it's x times y just write x y then you can have a quadratic expansion so we might have something like five x plus two and nine x minus three so again multiply the terms together five x times nine x could let the numbers first five times nine is forty five then look at the letters x times x is x squared so you can do them separately then we do five x times negative 3. look at the numbers first 5 times negative 3 is negative 15 and then we've got an x in there as well next we're going to do 2 times 9x which would be 18x and then 2 times negative 3 which is negative 6. so they are the four multiplications you're going to do the first term in the first bracket is multiplied by both things as i can break it then the second term in the first bracket is multiplied by both things in the second bracket now i'm just going to write the answers underneath a bit more space and what you'll find in quadratics is often the middle two terms can simplify negative 15x and a positive 18x so 18x take away 15x is 3x so our final answer would be 45 x squared plus 3x take away 6. and we can't combine the x terms the x squared and the x they stay separate and we can't combine things with x and normal numbers so we have three terms for our final answer all the things to look for uh quadratics with thirds involved so we could have x lots of root seven plus five all squared and the squared symbol means we're going to have this bracket twice and then we can do our multiplication so x root 7 times x root 7. so what's going to happen there is we're going to have the x times x which is x squared and then the root 7 times root 7 now if you multiply a root by itself it's going to cancel out the root symbol q squaring it so that'll be seven so it'll be seven x squared we'll do x root seven times five which will give us five x lots of root seven then we'll do five times seven so actually that's the same multiplication that gives them a lot of it then at the end we'll have five times five which is 25. now again it's a quadratic because often the middle terms will simplify you've got five watts of x37 and then another five watts of x root 7 so all together we've got 10 lots of it so our final answer is 7 x squared plus 10 lots of x multiplied by root 7 and then o25 one last thing to look out for is a cubic where you have three brackets so let's say we had x plus one x plus two and x plus three now the method here is to start it off like quadratic and only look at the first two so i'm gonna highlight the first two and multiply them out x times x is x squared x times 2 is 2x and 1 times x is 1x that's 3x all together and 1 times 2 is 2. and once you've got that then you can multiply that by the third bracket and again you're gonna use the same method you've just got three terms this time now i'm not gonna draw the arrows here i'm gonna do x squared times x is x cubed and then i'm gonna do x squared times three is three x squared so that's the first term ticked off then we do the second term 3x 3x times x is 3x squared and 3x times 3 is going to give us 9x that's our second term done now we do the third term 2 times x is 2x and 2 times 3 is 6. then all we need to do is add all of that together to get our final answer and all those terms together we've got our x cubed term only got one we've got two x squared terms three and three which is six x squared we've got two x terms nine x and two x which is eleven x and the number ten at the end the plus six so again if we see a cubic pretend it's a quadratic multiply the first two once you get that answer then write in your third bracket and pretend it's a quadratic again and multiply through again so it's basically a cubic is two quadratics in a row for quadratic equations you might have something like x squared minus 2x minus 24 and the task here is to put it into double brackets so we need to factorize this now we know that x times x will give us x squared we need to think about how to get the negative 24. now a good technique for this is to list the factors for 24 so you can divide it by one two three and four you can't divide it by five and while you can divide it by six we've all written that down so we've finished our list so just note that i went through all the numbers one by one to get a complete list of factors and then jump around randomly now everyone two numbers which multiply together to make 24 but they're going to add together to make a negative 2. so we have to look through our list and say well which one of these since they all make 24 which pair of numbers could make a two and it looks like it's going to be the one at the bottom four and six omega two and how does one six omega two you would have a negative six and a positive four to make a negative two so now it's factorized an additional question might be something like having it as an equation equal to zero and to figure out what x would be if it's all equal to zero you simply have to make x equal zero or rather one of the brackets equal zero so for example in the first bracket if x was equal to six the first bracket would read six take away six which is zero and that'll make the answer equal to zero because no matter what the second bracket would be and it would be two in this situation zero times two is zero in the same trick with the second bracket x would have to be negative four to make the second bracket be zero it would be negative four plus four which is zero and then no matter what the first bracket is i think that'll be negative ten because you're multiplying by zero you're going to get a zero you can also have quadratics like this one with a coefficient on the x squared so 3x squared minus 12x minus 50. so again we want to get double brackets you know what two numbers which multiply together to make negative 15 as the numbers in those brackets and the x terms well the only option is three x times x we'll make three x squared so let's have a look at the factors of fifteen so we could have uh one and fifteen we can't divide by two we divide by three that'd be three and five and we can't divide by four so we've finished our list so how can we get a 12 from those numbers well we can but another trick is because we have a coefficient on the x squared the three and that's throwing its way into one of the brackets that's actually going to multiply the second bracket by three and for multiplying the second bracket by three then we're allowed to multiply one of their factors by three so let's have a look at more options we could multiply the one by three or we could multiply the fifteen by three we can multiply the three by three or we could multiply the five by 3. now which of these options is going to give us a negative 12. it looks like um we've got a couple of answers here we could have 3 and 15 we've got 3 and 15 either way actually so what's going to happen 3 and 15 you have to think about where these numbers have originally come from so i'm not going to write 3 and 50 into the brackets i'm going to write the original numbers which the first option is 1 and 15. now because the 1 is multiplied by 3 that's going to be in the opposite bracket and so it would look like that then what a negative 12 so that'll be a negative 15 plus something to go up to 12. if we look at the other option and you can have multiple answers sometimes then the 3 and 15 is actually the 15 was multiplied from the five times a three so the five would have to be in the opposite bracket to the 3x to go up to 15 with the three staying as it is then again for this the largest one's gonna be negative so i'll be negative five and a positive three now interestingly you have multiple answers what is solving it equal to zero you will always get the same answer so for the first set of brackets three x plus three that means that three x would have to be negative three and if three x is negative three divide three by three x have to equal one now you'll also notice that the second bracket of the first option if x was negative one there that would also give you zero so two of the four potential brackets can give you negative one that means the other one would have to be a positive five for the second bracket the second option again the first bracket in the first option if x was five there that would also give zero now both of those examples have two answers for x but you can have one two or zero answers and the way you can figure out is using something called the discriminant the discriminant is b squared take away for a c so let's have a look at some examples so let's have a look first at the x squared minus 2x minus 24 that we stay started with a is the coefficient on the x squared in this case it's one if there's no number b is the coefficient on the x and c is the number at the end so b squared will be negative 2 squared take away 4 times a which is 1 times c which is negative 24. that's going to give 4 take away a negative 48 which altogether is going to give us 52. now what does this tell us about the number of roots well we'll leave that for a moment and come back to this let's have a look at another example so let's have a look at x squared plus x minus 8. it's good a b and c are going to be the same things a is the coefficient on the x squared b is the coefficient on the x and c is the coefficient at the number on the end so b squared would be 1 squared and we're taken away from that 4 times a which is 1 times c which is 8. all together that's going to give us 1 take away 4 times 1 times 8 that's going to give us 32. so all together it's going to be negative 31. okay so that's a negative number now again i'm going to explain what that means look at one more example then we'll explain all three together so we're gonna have a look at four x squared minus four x plus one so we have b squared that'll be the negative four squared again the term on the x taken away four times a which is four and c which is one that gives us 16 take away 16 which is zero and now we can talk about what all these mean with the discriminant so if you have a positive like in the first example let's write positive as if it's greater than zero that means therefore we are going to have two roots we have two answers for x if it's less than zero like in our second example that means we have no roots there is no way that you can factorize x squared plus x plus eight and finally if we are equal to zero then what that means is there's gonna be one root it's only gonna be one answer so when you have your factorization you've got your two brackets you'll find that both brackets have actually got exactly the same number in them and then that means that you only have one answer another formula is the quadratic formula and the quadratic formula is negative b plus or minus so you're going to do it both ways and get two answers the square root of b squared minus 4ac and you'll notice that's discriminant and that's why a negative discriminant means there's no answers because you've got to square root negative number you can't do it and that's all over 2a with the a b and c coming from the same source as they were for the discriminant a the coefficient on the x squared b the coefficient on the x and c is the number at the end now you can substitute into quadratic formula and that should give you your answers so if we have a look at an example let's do a first question again and see we get the same answers so for our first question negative b would be a negative negative two so we do have double negatives quite a lot here then we have the square root of b squared which is negative two squared minus four lots of a and six of four times one for a and times negative 24 for c all of that is going to be divided by two lots of a so that's two lots of one is two now negative negative two is fours with four plus or minus the square root of and we worked a discriminant for this already and it was 52 and then we divided everything by 2. and actually looking back that 52 was wrong because i multiplied by 2 and i should have multiplied by 4. so actually four times one times 24 would be uh 96 instead when we add the four to it that's gonna give us a hundred now a hundred makes more sense actually because then that is a square number and that means we can simplify this nicely so i knew this is wrong straight away because the root of 52 wouldn't actually give me whole number answers now you work this out 4 plus root 100 which is 10 is 14 divided by 2 which is 7. which isn't one of the original answers i've got so i know i've made a mistake somewhere either with the original question my work in here i'm just looking through things i've noticed negative negative 2 ended up being a positive 4. now actually negative negative 2 is just going to be a positive 2. so now it's going to make more sense because now i have 2 plus root 100 2 plus 10 which is 12 which divided by 2 is going to give us a 6 that we have in our final answer then once we try it with a plus we try with a minus 2 take away root 100 so 2 take away 10 gives us negative 8. we'll divide that by 2 and we get the negative 4 we had in our original answer as well now if you did this and then the root wasn't a square number and you'd be getting a decimal you rooted it then you've got two options you know the worker tote and have a rounded decimal answer or let's just leave your answer as 2 plus or minus root 100 over 2 all the equivalent of something that couldn't be simplified and that should be fine as well unless you've been asked to provide two roots or equivalent and you've maybe been given watch around it too we are going to go through a few different ways of solving simultaneous equations and by simultaneous what you mean is we're going to have two equations so we could have 2x minus 5y equals negative 14. and 5x minus 4y is equal to negative one so we have two equations the reason why we have two equations is that we can't work what x and y are unless we have two of them so we've got two variables we need two equations now the way to solve this is we need to make the coefficients of either x or y be equal and the only way to guarantee that is to multiply them by each other so let's say we're going to try and make the x's be equal to each other the two and the five multiply the two by the five and the five by the two and when you multiply make sure you multiply all the terms so two x times five is ten x and negative five y times five gives me negative 25 25y then negative 14 times 5 would be negative 70. then do the same thing with the second one this time we're times by 2. 5x times 2 is 10x negative 4y times 2 is negative 8y and negative one times two is negative two so what we've done now is that we've made the x terms be equal they've both got the same coefficient so now they're the same coefficient we need to cancel out to get zero and the way this time is by taking them away from each other 10x takeaway 10x is 0. sometimes you have to add and that's when the signs are different on them but in this case they're both positives we can take away so taking away the 10x away from each other 0 then negative 20y take away negative 8y all together it's going to give us negative 33 y or a little that way but i have made a mistake now if you look at this more carefully it's negative 25 like take away a negative eight so we've got a double negative so it's not actually negative 33 we're gonna be adding eight on which is going to give us negative 17y so now we're taking away two terms now check with the third term which is negative 70 take away negative two again that'll be double negative gives us negative 68. now negative 17y equals 68 that is an equation we can solve we can divide both sides by negative 17. so negative 68 divided by negative 17 gives us four so y is equal to four now that we know that y is equal to four we can find what x is and we do it by substituting into one of the original equations i'm going to choose the bottom one because i think it's got smaller numbers overall so it doesn't matter which one we choose the same answer right away so 5x minus 4y equals negative 1. so we know that y is 4. you can see that 5x take away 16 is equal to negative one and then we can solve that we can add 16 to both sides giving us that 5x is equal to 15 and divide both sides by 5 so x is equal to 3. the last thing you do is record your answers really clearly so x is equal to three and y is equal to four so we've got a few steps here we have equalized one of the variables either x or y either is fine they do that by multiplying them by each other then we've done an add our takeaway to cancel out that variable in this case it was a takeaway and we solved the equation then we substituted into one of the original equations for the number we just got y equals four and then we solve that so all together doing around about five things here other examples to be aware of are things like x squared plus three y equals fifteen so we've got a squared sign in this one then we have them without a square sign x plus y equals five so another way to solve these is slice substitution what we can do is we can rearrange this to make y the subject and we can say that y is equal to five minus x i can substitute that into our original equation so we've got x squared plus three lots of y and we know that y now is five minus x so if we expand the brackets we get x squared is equal to the x term first three times negative x is negative 3x then the number term 3 times 5 is 15. and then we can uh take 15 away from both sides to get it equal to zero and if we do that we're actually going to get it equal to zero without a number term on the left hand side now we have a quadratic and by working out the quadratic we're going to get our answers now if factorize this quadratic because there's no number term at the end we're just going to take x8 so i divided both terms through by x so to find out what x is we want it all to equal zero so if x was zero that would give us zero times what's in the bracket which is equal to zero also if x is equal to three then the bracket itself be three take away three is zero and then you multiply that x by zero so there's two different ways to make this equal to zero all we do now is substitute those numbers in to our original equation that would give us zero plus y is equal to five and it would give us three plus y is equal to five that means that y would have to equal five r y would have to equal two at the end writing final answers our first set of answers was x being equal to zero and y being equal to five and our second set of answers was x being equal to 3 and y being equal to 2. the last thing to be aware of is that you can do this as well with kind of line equations so for example if we said y is equal to 2x minus seven and y is equal to the square root of negative two x plus nine what we could do with these two is that we could combine them together they're both equal to y so we made them equal to each other and that'll give us two x minus seven is equal to the square root of negative two x plus nine and then recall to each other you can rearrange them and then that's going to give you a quadratic because our first step is going to be to root both sides up right square both sides to get rid of the root that would give us 2x minus 7 squared is equal to negative 2x plus 9. now i'm not gonna do the rest of this because we've seen this already but we rearrange it get a quadratic and it's like the last example you're probably going to get two answers for x and once you get two answers for x and substitute into any original equations to figure out what y is we are going to have a look at how to factorize a cubic and the one we'll look at is x cubed minus x squared minus eight x plus 12 and so what we want from this is we're going to run three brackets that all multiply together and they're going to make this number so for example for the x cubed we know that must be x times x times x but then for the extra numbers we know that the numbers multiplied together must be the 12 but there's different ways of making the 12. we could have 6 times 2 times 1 we could have 3 times 2 times 2. there's lots of different ways of making it and we don't really want to use trial and error and expand out lots and lots of brackets so we need to use a simpler method now we are going to use a little bit of trial and error and the way we do this is using substitution i need to guess what our first factor is going to be you'll see why in a moment so let's guess that our first factor is gonna be when x is equal to one if we substitute that in we get one cubed take away one squared take equate eight lots of one plus twelve now if that equals zero then we have a fact so we have one take away one take away eight plus twelve all together that actually gives us four that means that one is not a factor so now we could try negative one or we could try uh positive two so that'll be for two two cubes take away two squared take away eight lots of two plus twelve that will give us eight take away four take away sixteen plus twelve and all together that does give us zero that means that two is going to be one of the answers and so if x is two then we're going to have x minus two in one of the brackets because i'm gonna say it's equal to zero if this whole thing was equal to zero then x being equal to two would mean that our first bracket is zero because two take away two which makes the whole thing equal to zero so now we know one of our factors now we can use the method to find the other two factors and that method is algebraic long division so we're going to take our factor x minus 2 and we're going to divide our expression by it that's x cubed take away x squared take away 8x plus 12. so we use algebraic long division so the way this works is we can't really divide through by x minus two that's a bit too complicated we're only going to divide by the first term the x i'm going to do each term one at a time look at x cubed first so let's divide x cubed divided by x is x squared now that's not actually our answer we didn't divide by the minus 2. so what we do now is we multiply the x squared by our original uh divider x squared times x is x cubed and then x squared times negative 2 is negative 2 x squared so what that means is we didn't actually get the negative 2x squared out of this so now we can include it so by doing that division we've accounted for per the expression x cubed take away 2x squared so we take it away you know from the full expression now the first two terms will calculate to make zero then we have negative x squared take away a negative two x squared we could take away double negative like that that'll be negative x squared plus which would be a positive x squared and the rest of the expression is having zero taken away from it so it stays the same so we divided by our first term and we've seen what's left over now we're going to divide our brand new second term not the original second term and new one by x so x squared divided by x will give us one x then of course we didn't divide by the minus two so we multiply it to account for the minus two x times x x squared x times negative two is negative two x then we take what we found away from what's left over of our expression first two terms can't have to pick zero and a negative eight x take away negative two x double negatives what not that shows a negative six x and then we've got the plus twelve that we've taken nothing away from so that's our second term dealt with now we look at the third term now it's the third term from our new expression negative six x and we're dividing it through by the x so negative six x divided by x is negative six then again we didn't divide by the negative 2 so we've got to account for that by multiplying by it negative 6 times x is negative 6x and negative 6 times negative 2 is a positive 12. if we take those away now we get 0. and if you get zero you have no remainder that means that x minus two does actually divide into x cubed minus x squared minus eight x plus twelve and therefore it is a factor so it's kinda proven what our substitution showed us earlier on now what that leaves us with is this we know that x minus 2 is a factor but also the answer to the division of the factor x squared plus x minus six and now what we can do is our first bracket's fine but our second brackets are quadratic and we can turn that into two kind of linear brackets so now you just use your factorizing quadratic skills now negative six want two numbers which multiply together to make six and they're going to add together to make a one x well two times three would make six and then to get a positive one and do positive three take away two so we're just using our factorizing quadratics using inspection skills and there's a factorized cubic is x minus 2 x minus 2 and an x plus 3. we can even simplify that we've got two lots of the first bracket we can give a little square sign there and then write our third bracket afterwards functions is something we can use for substitution for example so we have the function which we call f of x a function of x and we say it's equal to three x plus five then if we want the function of let's say five then we swap all instances of x with five that'll be three lots of five plus five three times five is fifteen and plus five is twenty so that's the primary use of functions now there's some things with functions that you need to be able to do so the first thing to do is substituting into functions multiple times so let's say we want to find f g of 3. well what could that mean well if we say that g of x is equal to 11x and f of x is equal to x plus three now we've got two functions that we can look at so we've got the function of x x plus three and for g of x it's not f function but we can't use the same letter or everything so it's just another function 11x so to find gf of 3 what we do is we start with the inside we start the letter that's nearest the number so i'm choosing which is the f and so what i'm going to do first is work out f of x and that's gonna give me f of three which is three plus three which is six another f next we do g and it's gonna be g of our new number and new answer which is six and g of six that's gonna be eleven times 6 which is 66 for our answer so to substitute into multiple expressions we start off with the expression that's nearest the number substituting we'll do f first and then we do g now we can also find an expression that kind of is is g of f of x we can combine them together without doing substitution now the way we do that is we're going to start off with the outside with g and g is 11 x but what we're going to do is we're going to swap the x with the function f which is x plus 3. now if we substitute into that and so we shoot 3 in we should still get 66 so let's test it so g f of 3. that'll be 11 lops with 3 plus 3 3 plus 3 is 6 and 6 times 11 is 66 so that method does actually work so we've looked at just regular functions and substituting in and looked composite functions where we combine multiple functions together the next thing to look at is inverse functions so let's say we have a function of x and it's x minus nine all over six now we can find the opposite of this so we put a number into this and got an answer let's say x was 10 then we'd have 10 take with nine is one and one divided by six is one sixth so we get one sip now the inverse function is a function where we put in one sixth and we get ten out the other side we work the opposite direction now the method to do this is firstly translate this into a kind of line equation almost so y is equal to x minus 9 over 6. and then once you've done that flip around the two symbols so x and y are going to be the other way around so that will give us x is equal to y minus nine over six and now it's inverse let's rearrange it to make y with subject so we can multiply both sides by six that will give us 6x is equal to y minus 9 and then we can add 9 to both sides and that will give us 6x plus 9 is equal to y and once we get to that stage all we do now is we swap y back for f of x but this time f to minus one x showing it's an inverse function which is what the negative one does the inverse function is six x plus nine so remember example earlier if we substitute one sixth into this we should get ten out of the other side so if x was 1 6. we'd have 6 lots of 1 6 which is 6 over 6 which is a whole one and 1 plus 9 is 10. so it checks out that's actually going to work so let's run through the kind of things we've done we've done substitution with functions we've drawn composite functions we combine multiple functions together and we have done inverse functions so three different types partial fractions is about taking an expression let's say 12x minus 8 all divided by x squared minus 5x minus 24 and changing it rather than being one fraction changing it into being two fractions now the way you change it into two fractions is all about how you factorize this so your first job is to factorize now if factorizing the numerator you can take 4 out as a factor and factoring the denominator it's a quadratic so you can put it into double brackets and we should have x plus 3 and x minus eight when you factorize the quadratic using whatever method you like now once you have that now we can see the fractions we need again it's still one fraction our new fractions are gonna have x plus three and x minus eight as denominators but then the numerators we don't know what they are going to be so let's call them a and b now we're going to have to try and simplify that and so what i'm going to do is i'm going to write everything out again so i've got some working out space because i'm going to do a drawing all over this and doing lots of crossing outs all over this so what the left hand side written out again we also want the right hand side real nice again and what we're going to do is try and get rid of the fractions that's our first step so what i want to do is multiply through by all the denominators so we have x plus 3 as one of the denominators so we're going to multiply all three terms we have here by x plus three then the other denominator is x minus eight so we're gonna multiply everything by x minus eight as well now you might think what's the point with this because we haven't got rid of the denominators but then we can simplify three we can say well x plus three divided by x plus three is one and so they can't slightly multiply the numerator by one it's going to stay the same and x minus a divided by x minus a that will cancel out as well do the same thing on the right hand side we've got x plus 3 and x 3 cancel out and x minus a and x minus 8 comes slope so only two of the brackets that are multiplied through by are still going to stick around and what that's going to give us if we then write this out again is 4 lots of 3x minus 2 and that's all that's left on the left hand side and on the right hand side we'll have a lot of x minus a plus b lots of x plus three now we can substitute in values of x to get rid of one of the variables because we can't solve equations got an a and a b we need to get rid of one of them so we substitute in two things to get rid of a x would have to equal eight because then the bracket would be eight take away zero and a times zero is zero so if we said that x was equal to eight that would give us four lots of three times eight take away 2 on the left hand side and on the right hand side the a bit's gone so we've got b lots of a plus 3 and then you can wear that out so on the left hand side 3 to 8 is 24 then we take away 2 to get 22 and we multiply by 4 to get 88. that's equal to 8 plus 3 is 11 so that's 11b so b must be equal to 8 so this is when we got rid of a and we got b now i need to get rid of b to get a now in the b bracket we have a positive 3 so if we took away 3 from that negative 3 plus 3 is 0 and b times 0 is 0. so let's substitute that we have 4 lots of 3 times negative 3 then i'm going to take away 2 and that's going to be equal to a lots of negative 3 take away 8 and the b bit has cancelled out to give us 0. working this out 3 times negative 3 is negative 9 take away 2 is 11 and multiply by 4 is negative 44. right hand side negative 3 take away 8 negative 11 that's negative 11a so divide through by negative 11 and a is equal to four so now when we go back to writing our fractions let's write an answer a is four so that would be four over x plus three and b is eight so that would be eight over x minus eight so i'm just gonna shade that in now to show it's our final answer so it's highlighted in pink this is our answer now there's a few extra things here so for example let's say that b was negative eight what we would do is we would have plus negative eight and negative eight as a numerator what you would do is have a minus eight so you can bring the negative symbols on the numerator down to be the kind of symbol between the fractions the other thing is this is a regular fraction the numerator is smaller than the denominator and very small we're looking at the powers on the x you know 12x we've got x to the power 1 there then with the x squared that's x to the power of 2. now if that was the other way around and you had a larger kind of value of x on top in terms of its power if the power of x was larger on top then what you would have to do is divide it to try and turn it into a mixed number and the way you do that is algebraic long division and if you divide the two things you can also create long division then you'll get a brand new fraction that you could turn in we're gonna plot first the line y is equal to negative two x minus three so there's two key parts to this the first part is the y-intercept and the y-intercept with the line is going to cross the y-axis it's going across at negative three so it's a full quantity zero negative three it's got zero as the x-coordinate because we cross over the y axis that's where x is equal to zero the second important feature is the gradient which in this case is negative two so with the gradient the gradient is a change in y over the change in x now if it's a whole number we need to write this as a fraction what's it y over x so i had your whole number well you can write whole numbers as fractions you just write them over one so what we're saying is every time the y coordinate changes by negative two the x coordinate will change by one so if you go down on the y coordinates and across one of the x coordinates the place that was highlighted in pink there we've gone down by two for the y direction and across by one on the x direction and we just keep repeating that keep repeating that pattern over and over again and also do it in the other direction the opposite this is going up two and then going across negative one and we get a line like this now i'm just going to sketch this line you should be using a ruler and a pencil i just want to show you to join up all the points so what we have here is the line y is equal to negative 2x minus 3. now something else we'll look at here what if we're not plotting that at all what if we're plotting the modulus of negative 2x minus three and for the modulus we have these they're almost like brackets but they're just kind of straight lines at either side so what does this mean now with the modulus that means that we can only have positive values so let's write some notes on that the modulus so the lines either side so it means positive only and on the same positive what we want actually is positive um y values first so we look at our line you'll notice that the first part and left hand side all of this is positive already all the y coordinates are positive the x coordinates are negative so it's only the y coordinates that we're worried about with them being positive now we've got the positive side it then becomes negative and what we're saying is the modulus only has positive y values so when we go down to the next coordinate we've got plotted here which is negative one negative one what we actually want to put is negative one positive one so the y quad is now positive with the y-intercept which is zero negative three then it'll be zero positive three the next one along that i plotted was one on the x axis and negative five on the y axis so one negative five becomes one five and see the same kind of gradient pattern here but it's going in the opposite direction and so then when you're plotting the modulus you're going to plot this section here and join it up and then this is the line y is equal to the modulus of 2x minus 3. now the effect of the modulus is that you can probably see it's kind of almost like the line's bouncing off the x-axis so with our original line it will go straight through the x-axis but the modulus is now bouncing off so that's something to be aware of being asked to find the modulus of something the next thing we're going to look at is we'll look at solving equations using a graph so we're going to solve 3x plus 8 is equal to negative 7x take away 2. and we can solve this using the graph now the way we do this is first we have a look at the left hand side we've got three x plus eight we are going to draw the line y is equal to three x plus eight then we look at the right hand side and we've got negative seven x take away two so we're gonna draw that y is equal to negative seven x minus two now the green and the pink part are both equal to each other in our original equation so we can make them both equal to y they're still equal to each other so let's plot the two lines so three x plus eight the y intercept is at eight and the gradient difference in y over x is gonna be three over one so when we go up three we're gonna go across one but that's gonna go off the graph actually so let's go in the opposite direction we'll go down by three we're going to go across by one and so we can plot the graph like this then we can draw a straight line through so there's our this line now look at our second line negative 7x take away 2. so the y intercepts at negative 2 and the gradient is going to be negative 7 over 1 so we go down by 7 on the y axis we'll go across by one on the x axis now we're going to go down once as you can see before we run off the edge so let's go in the other opposite direction we're going to go up 7 and i'm going to end up like that and i can't really plot the next one now the important thing here is i just plot the little line segment i've got i can estimate what it's going to carry on so if you use a ruler there you're going to get a really accurate line the important thing about these two lines is you can see that they both cross over each other they both intersect so we want to look at the point of intersection now the x-coordinate is negative one and the y-coordinate is at five and this is actually the solution to our equation because we've got an x and a y coordinate and we're looking for x's and x is negative one so that's our answer x is equal to negative one so rather than doing this algebraically you can do it on a graph just by plotting the left and right hand sides separately make them both equal to y and the place where the lines crossed over is your answer next we'll plot quadratic lines so our first one is going to be y is equal to x squared plus x minus two now the usual method of plot things like this is to substitute so if you choose a coordinate and substitute in you're gonna get some values you can use so let's say that x is equal to three when x is equal to 3 that's going to give us 3 squared plus 3 minus 2 which all together is going to be 9 plus 3 minus 2 which would be 10. and what you just found there is the y coordinate so when x is 3 we'd have the coordinate 3 10 and we can plot that on the graph now we could do that for every single value of x but this is going to take a while so let's look at a few shortcuts the first shortcut is that the number on the end of this the negative 2 that's the y-intercept just like with straight lines so we know that our curve is going to go through negative 2. now the y-intercept is when x is equal to 0. this means we don't need to substitute in x for 0. we can also factorize this i'm going to go through it to factorize we've got a revision slide on that when you factorize this you're gonna get x plus two and x minus one now if we solve this for being equal to zero we're gonna get that x is equal to negative 2 and x is equal to a positive 1. and so it's the opposite sign because what we're trying to achieve here to get one of the brackets to equal zero so we multiply by zero and get the answer zero so it's equal to zero so by solving the quadratic we know the x intercepts we know we're going to go through x is one and x is negative two so i've got two more coordinates again without having to do any substitution now once you've got these you can then have to think about well what else do i need to substitute so we can see we've got negative two zero one three we don't have two so it might be you might wanna substitute two in and see what that gives us so two squared plus 2 minus 2. that's going to give us 4. so that'll be the coordinate 2 4 and we can plot 2 4. so it's almost like we don't really want to substitute for every single value so we can use the x and y intercepts to give us an idea of which one to substitute because then after x is three if we substitute x equal to four it's going to go off the graph and it'll be a value that we can't use so we don't want to substitute in anything larger than those there's one more thing that's useful and this is by differentiating this we can find the turning point of the curve so let's differentiate d y by dx we bring down the power x squared becomes 2x and then when you bring down the power you use it by one so power 2 becomes power one x is to the power of one so you bring down the one and then if it's power one it's gonna go into power zero and disappear and then the negative two doesn't have any power so we're gonna ignore that if basically power zero and you multiply it by zero it's gonna go away so d y by d x is two x plus one what does this help us well that's the gradient of the curve so when that equals zero that's when the curve has a straight gradient which is going to be the turning point so we say that two x plus one is equal to zero for the turning point that means we've got two x is equal to negative one so x is equal to a negative half that gives us the coordinate for the turning point well we know that x is going to be negative 0.5 you'll see we don't have the y coordinate for that so we have to substitute it in so that would give us negative a half squared plus negative a half take away two so a negative half squared is going to give us one quarter e quarter plus a negative half so plus negative it's going to calculate to be a negative take away so we're going to have a quarter take away a half we'll give us a negative quarter and then we take away two so it's going to be negative two and a quarter altogether that gives us the coordinate negative half negative two and a quarter and that's going to be the turning point of the curve so with that a little tricky to find roughly here we can see the turning point actually is not too far below the y intercept for this one now you'll notice there are some missing coordinates and the trick here is that these graphs are symmetrical so for example on the opposite side the turning point is going to come up to the same height as the y intercept then we've got the x intercepts which we both have and then on the right hand side after the x and set it goes up by four so on the left-hand side is also going to go up by four it's going to be the same height as with the green coordinate then we go up to the top you go one to the left it's going to be the same height as the next coordinate so you can use a symmetry of the curve to fill in the missing coordinates and now we can sketch it and don't do straight lines here try and make it curved you have to draw it freehand make sure it goes through all the points and there's the curve so let's review the techniques that we used we could find the y-intercept just by inspecting the formula we could find the x-intercept by factorizing and solving it for zero we could find the turning point by differentiating it and then we could find all the coordinates that we wanted by using substitution and you'll need to use two extra substitutions although the differentiation also requires a substitution to find the y coordinate of the turning point so that's how we plot a quadratic let's have a look at another example so we have y is equal to negative x squared plus three x so again we can find some things about this without doing any substitution we can look at the plus number at the end and this one doesn't have a number added on at the end that means the y intercept is going to be zero then we can factorize this to find the x intercepts now if you factorize this we'll take x out as a factor that's going to give us we'll take the negative out as a factor as well so it's a negative x over negative x squared we're left with x and then we take negative x out of 3x that's going to be negative 3. and if you solve that equal to 0 that means in the bracket x have to be equal to three and outside the bracket x has to be equal to zero so the x and sets are at zero and positive three it's interesting to this one that if you don't have that kind of number at the end and the y intercept zero it can be the case that the x and the y-intercepts are going to be the same in fact they will be one of them will have to be zero next we can differentiate this for the turning point so d y by d x is going to be bring the two down negative two x to the power one reduce the power and it's going to be plus three and what we want is for that to equal zero for the turning point so we can take three away from both sides and divide both sides by negative two so x is gonna be equal to three over two it's gonna be equal to one and a half so now we can substitute in one and a half for the y coordinate so we want uh negative one and a half squared plus three lots of one and a half so you could write three over two or you can write a one and one half it doesn't matter which way you write it so now we can substitute so we have the one and a half so i'm going to square it i want the negative version of that it's going to be negative 2.25 then we're adding on three lots of 1.5 which is going to give us 2.25 or two and a quarter the really important thing here is this is a negative half at the start don't square negative a half square the number you are substituting in which is the 1.5 one and a half and then pop the negative symbol afterwards remember you're doing disease first then use attractions so this will give us the coordinate one and a half two and a quarter which we can plot so that's the turning point of the curve now we need some more information to plot the bottom part of this graph so what we're going to need to do now is do some more substitutions now given that we've got values for zero uh one and a half and three it might be useful to look at four now i've got a space to write this on the page but we're gonna do four squared which is sixteen or the negative of that to negative 16 then we're gonna add on three lots of four that's we give us negative four all together so that's going to be the coordinate four negative four with the symmetry we also know that negative one's always gonna be d at negative four it's going to be the same height we might want to try uh 5 as well so we want 5 squares 25 so negative 25 plus 3 lots of 5 and that gives us negative 10. so we've got 5 across we've got 10 down and then negative 2 from the symmetry will be there as well so i'll start substituting just like i did with the one and a half and just like i did for the two and the three in the previous question so now we can draw the graph because i don't need any multiplication i can see that negative 2 and 5 have got me right down to the bottom so any of the substitutions will be a waste of time and so there's our final answer next we're going to plot a cubic we're going to plot y is equal to x cubed minus three x squared minus x plus three so again we can inspect this to find some information so we have a plus three at the end that must be the y-intercept going through three so the y-intercept is gonna be zero 3. then the y intercepts always have an x coordinate of 0. again like the quadratics factorizing this is going to give us some points to look at as well so if we take a full cubic again i'm not going to go through factorizing a cubic because we've got a revision slide on that so i'm just going to give the answer to it but remember some algebraic long division for this is going to be helpful so the factorization is going to be x plus 1 x minus 1 and x minus 3. again that factorization will take you a bit of time and that'll be worth you know a couple of marks on its own now if we solve it equal to zero we're going to get the x intercepts and to make this equal to zero one of the brackets has to be zero so either the first bracket would be negative one because negative one plus one is zero the second bracket could be positive one and the third bracket could be a positive three so they are our x intercepts we have negative one positive one and positive three now we've got these pieces of information about what the turning points of the curves and this can be done by differentiation so again we start off with a full cubic you differentiate it bring down the power and reduce the power by one you know bring down the power two and there's number one reverse you multiply 2 times 3 is 6 and reduce the power by 1 and you should get this now at the turning point this is going to equal 0 because that's where the gradient is flat there's no steepness at a turning point now if we factorize this quadratic we're going to get our turning points now again we've got a different slide for revising factorizing i think this was the user quadratic formula and so what it's going to give you is x is equal to 2.2 and negative 0.2 to one decimal place so we know the location of the turning points but we only have the x-coordinates we want the y-coordinates so to get the y-coordinates we're going to substitute those into our original equation so we're going to do 2.2 cubed take away 3 lots of 2.2 squared take away 2.2 and then add 3. and so for the y coordinate that's going to give us negative 3.1 to our decimal place for the first one then do the same with the second one we've got negative 0.2 we're going to cube it we're going to take away three lots of negative 0.2 we're going to take away negative 0.2 that'll be double negative so we add on 0.2 and then we're gonna add on three and that gives us three point one to one decimal place so now we can plot the turning points the first term point x is gonna be 2.2 and y is gonna be negative three point one and you can't plot it exactly with the kind of level detail let's just do the best you can the second turning point x is negative 0.2 and y is a positive 3.1 and actually it's really really close to the y-intercept so what we can do now is we can kind of plot the inside part of this curve like this what we need to do now is to see how this goes off the graph so we need some more substitution so i'm not going to write this down but basically you know i've got a point for 1 and then i've got just over two and i've got three so let's have a look at four so we'll have four cubes take away three lots of four squares take away four and plus three and that gives us 15. so what this tells us this is going to go off this graph really really quickly actually it's gonna be something like this it's not actually gonna make x equals four before it goes off so the top is going to be at y equals 10 when i get to 15 i'm going to be about two thirds of the way across there let's look at negative two so we want negative two cubed take away three lots of negative two take away negative two double negatives by adding two and then we're going to add three and that gives us negative fifteen it's gonna be the same thing in the other direction yeah by the time you get to 10 it's gonna be at two-thirds of the way across to negative two because when we get to 15 we're going to be the whole way across and so there's our cubic plotted uh pretty accurately i think actually so let's look at all the different techniques look at the y-intercept for the uh the number at the end of your equation you get the x-intercept by factorizing the cubic so you might be you know doing a bit trial and error here you know testing cancel due to one like give us zero you might be using algebraic long division and then we use differentiation to find the turning points once you've got those then what you might want to do is just try and substitute in a couple of extra points to get the full picture of what you're missing inequalities can be solved on a graph and some really good advantages to it you'll see in a moment so the first one we'll look at is y is greater than negative two x minus seven and y is greater than or equal to seven x plus two so what you want to do is to plot the two lines and let's start with the top line it's got a y-intercept at negative seven so we can plot that and the gradient is negative two so it could go down by two each time we go across one on the x-axis and then in the opposite direction it's gonna go up by two and we can draw the line now you'll notice i'm gonna draw a dashed line for this and the reason is this is a greater than line so we do a dashed line if it's greater than or equal to you do a solid line and we'll talk about the importance of this in a moment once we've got both lines our second line is y is greater than seven x plus two so we have our y intercept at plus two and our gradient is seven just gonna go up by seven each time in the opposite direction it will go down by seven each time and so we're gonna have something a little bit like this now you'll notice this one i'm doing a bold line a solid line this is closely greater than or equal to so you've got the equal sign involved then that's where you have your solid line now the benefit of this is that we're not looking at the point of interception here that would be solving the equation if these were equal to each other and we have no equal signs so we're looking at areas for our answer so let's look at the first y is greater than negative two x minus seven that means we want things that are greater than the line we want things that are above the line and you can see in blue i'm just highlighting above the green line for the pink line that's also a greater than so we want things above the pink line so we'll highlight that and what we want is you want to shade in the part of the graph that i colored in twice and that's a bit at the top you can see that's part of the green area and this part of the pink area so let's completely shade that in now the bits here towards the bottom on the left hand side that was only above the pink line on the right hand side it was only above the green line so we don't want those so we'll rub out the extra shading that i add to working out and so the shaded blue area is my answer that's where both of these inequalities are true now what you can do now is you can choose coordinates of that area and they'll satisfy both lines so let's say we choose the coordinate here which is um negative 2 4. so if you substituted in negative 2 and 4 what you'll find is switching to negative 2 for x you'll always be greater than four that's what it means and so you can have any coordinate in here as you answer but you will often you ask for you know exactly any coordinate now what about the dotted line and the thick line now with the thick line you're allowed to have coordinates that are on the line with the green line you are not allowed coordinates on the line so it's greater than or not equal to your load coordinates on the line any quadrant on the pink line is fine if it's just greater than that you're not allowed any of the coordinates directly on top of the line so that's something really important to take into account let's have a look at another inequality so this time we'll look at y is greater than negative x squared plus two x plus two i'm going to look at y is less than two plot the two lines so y is less than two plot y is equal to two but it's going to be a dashed line because it's not equal if we read a it's just less than it's not less than or equal to so we're not allowed any of the coordinates actually on that line it's only ones below the line then we're going to plot the quadratic so use all the things that we revised for plotting quadratics i'm not going to go through it in too much detail so we know it's going to go through to on the y-axis we know that if we factorize it we're going to get the x-intercepts and i'm going to go through factorizing but i believe it's going to be at 1 negative 1 and 2. and you can you know plot the quadratic mu substitutions and get accurate coordinates now i'm not going to plot this too accurately i just want you to get the idea of what we're doing and it's greater than so it's going to be again a dashed line so it's going to be something like this so then we solve this inequality again it makes you put the proper effort into sketching it make it you want to find more coordinates than i did there but y is less than 2 so we want everything below the pink line and y is greater than the quadratic so we want everything that is above the green line i want to shade in the area shaded in twice now the bit in the middle that's only been shaded in once so let's rub that out the bit on top that's only above the green line so i can shade that out yeah of left and right sides that's not only below the pink line but it's also above the green line so we can have the entire area on the left and then it's actually the same thing on the right as well so all of the bit on the right is below the pink line it's less than but it's above the green line is more than that and so we can shade in the entire right side as well as long as below the pink line and so for this question we've actually got two shaded areas and that's something that can happen with aquatics but not all the time sometimes you have a bit in the middle or the bit on top so you have different options but this is an out can we get sometimes with quadratics we can also solve inequalities algebraically so based example would be something that looks like an equation to solve so let's say two lots of eight x minus six is greater than or equal to sixty-eight just pretend you solve an equation but you keep the sign as it is rather than having an equal sign so you can divide through by 2 giving you 8x minus 6 is greater than or equal to 34. you can add 6 to both sides giving 8x is greater than equal to 40. and then you can divide both sides by 8 saying that x is greater than or equal to five so you just solve it like you would any normal equation you just have a different symbol now where this changes a little bit is when we have quadratics involved so let's have a look at x squared plus two 2x minus 8 is greater than 0. now if we pretended this was equal to 0 we could factorize this quadratic and so we factorize it x times x is x squared at negative 8 we need probably 2 and negative four and does that give us a positive two so negative four plus two would give us a negative two so actually we need a positive four and negative two and then a positive four t equation gives us two x and negative two times four gives us negative 8. if we solve the e equal to 0 we would have x is equal to 2 and x is equal to negative 4. however we don't have it equal to 0 it's greater than 0. now you might think well okay we can just put in the greater than symbol x is greater than two and x is greater than negative four what happens here is that the greater than negative the greater than two is a redundant because anything that's greater than two is also greater than negative four so it's a little bit of a clue that something's going wrong here now the key to this even though we're solving algebraically is we are going to need a little bit of a graph to show this so we draw our curve now i'm not going to draw it accurately now i know that the y intercept will be negative 8 and all the x-intercepts are right two and negative four and this is pretty much all the information we need really so i'm gonna sketch a curve not worried too much how accurate is you'll see why in a moment so it's gonna be something like this i was saying it's greater than zero so we also want y equals zero in here as well now just like when we're solving these graphically we'll do the same thing so we want greater than zero so when shading the parts of the curve are greater than zero and the parts are greater than zero are these bits here this is the part of the curve that's more than zero and so we'll look at this in more detail so the first bit that's greater than zero is happening at negative two and the second bit is greater than 0 is happening to be negative 4. now just look at where the graph is going we're going below negative 4 we're going towards negative 5. so that means that x is going to be less than the negative 4 because answers less than that are on the line on the other side we see the two we can see our lines going towards three that's gonna be things that are greater than two and so we have two different answers with different symbols different arrows playing in different directions x can be greater than two that's above zero or x could be less than negative four because then the graph is also going to be above zero in terms of the y coordinates so again we've got two answers because we've got two separate parts of the graph shaded in now let's have a look at another example so we're going to have a look at negative x squared plus 7x is greater than or equal to zero so again we want to square sketch the quadratic now there's no plus number at the end so it's going to be zero for the y-intercept if we factorize this we're going to get something like this so if we pretend that it's equal to zero x would equal zero and x would equal a positive seven so we know that seven is going to be an x coordinate as well now if we sketch this graph it's going to look something like this and again i'm not too concerned accurate this is i don't know how high it's going to go actually but it's going to be something like this and we want when it's greater than or equal to zero so which part of the curve is above zero for the y-axis to shade it in it's the kind of loop in the middle now let's have a look at the numbers so we've got where x is zero and we've got an x is seven so you can see that we're allowed values in between zero and seven so then when we're writing this down we know x equals zero and x equals seven we want x is going to be greater than zero and x is going to be less than seven and then because the originals greater than or equal to we can pop the equal signs on those as well another thing we can do is we'll write that in an alternate form so x is greater than or equal to zero and x is less than or equal to seven they can be combined into zero is less than or equal to x and x is less than or equal to seven so both of these are the same thing they represent the same information but you can just combine them together so you write down less x's left to right basically now have a really careful look at our two examples here so what we've just done above zero on the y-axis you can see we've got the kind of curve and it's all in one piece in our first example we can see when y is above zero that we have two separate pieces so that's just a clue to how to write it two separate pieces you want two separate answers like x is greater than two and x is less than negative four if the whole thing's together and that's the clue you want it all combined so you want zero is less than equal to x which is less than equal to seven looking at transformations of functions we've got the black line on each graph and this is the line f of x so we don't actually know what the formula of this line is now the quadratic is going to be an x squared we don't know exactly what form that's going to take so each one is y is equal to f of x so let's have a look at ways we can transform this so on the first graph in blue we're going to do y is equal to f of x plus 2. and so if we do that what's going to happen is the graph is going to shift two places on the x-axis because 2 is next to the x and so if you just take important coordinates for example the x intercepts these are going to shift two places across and so the base of this graph the curve is going to be here and then where it goes off the graph up the top that's going to shift across two places so roughly looking at a graph like this now you might think well this has gone backwards by two it's not gone forwards by two and so you're always going to think when you're transforming functions it's gonna do the opposite of what you think you're almost saying if you add an x to two you want x to appear two places sooner if you take x to be say time and from left to right you're going to follow through time so you've added two it's appearing two blocks sooner and so we can say about this if you look at the coordinates each coordinate has gone down by two and the y coordinate has stayed the same let's have a look at another transformation so let's have a look at y is equal to f of x plus two now this time the plus two is on the outside of the brackets so the effect you can rearrange this and if you take two away from both sides but y take away two is equal to f of x and so we can see that around the last one where the x had something added to it this time the y is having something added to it you're going to do the opposite of what you think so if y is minus 2 you're actually going to add two so again take important coordinates like for example the x intercepts they are going to go up by two for a new graph where we go off the graph that i got by two now that's going off the graph so it might not be the most helpful to look at those you could look at the y-intercept that's gonna go up by two as well they see the y-intercept is one across from the x-intercept so it's gonna be one across another direction as well so now we can do a little bit of a sketch of this line so it's going to look something like this so the whole thing has shifted up two places now again if you look at the coordinates what's happening to the quadrats is the x coordinates are staying the same but the y coordinates you're adding two to them so what we have here is these are basically translations so we have translation on the x-axis when you have your number inside the brackets with the x and then we have translations on the y-axis with the numbers outside the brackets and if you rearrange it you can see it'll be next to the y so let's have a look at some more transformations so let's move on to the second graph because i want to write too many on the same graph because it's going to get confusing so the next transformation is going to be y is equal to f of 2 x so we think about what's gonna happen with the coordinates here it's on the inside of the brackets the numbers next to the x so we're going to have something happening to the x coordinates it was the opposite of what we expect this one's being multiplied by two so the opposite of that is going to be divided by two so the x coordinates it could be divided by two and the y coordinates will stay the same so again find some important coordinates so again we got the x intercepts one of them is at negative one so divided by two it'll be negative half one is at negative three divide that by two and it's gonna be negative one point five so we know the curve is going to be here the y intercept is when x is zero so that's not going to change if you divide zero by two you still get zero and you use the symmetry to have the point on either side as well so what about were the graphicals off the end of the page well one's just after one so that'll be half of it just after 0.5 and then the other one is at just over negative 5 so divide that by 2 and just over negative 2.5 and then we can draw that and so what you can see is if you divide the x coordinates by two you actually make the curve kind of more squashed together it takes up half as much space on the x-axis let's have a look at another so we're going to look at y is equal to 2 lots of f of x now again this we can rearrange it to get that number onto y so that would be y is equal half of y is equal to f of x so it's divided by two now as usual it does the opposite of what you'd expect so if you're dividing the y coordinate by two in the equation in our answer we're gonna be multiplying it by two this will be y times two so we multiply all the y coordinates by two now the x intercept's really important here because that's been y zero if you multiply zero by two you still get zero that's not going to change places what will change where it goes off the page then that's where y is 10 so that's going to end up at y is 20. well we can't plot that but we'll look at the y is 5 when we double that we'll get 10 which is where it goes off the page now all you have to do is look at the two places where y is equal to 5. if you double it to 10 you just draw the same coordinate at the same place on the x-axis but then um moved up to 10 on the y-axis and we can actually uh probably draw this now perhaps one more important point is the turning point of the curve so that when y is equal to one so if you double it it'll be at y is equal to two and so the curve is gonna look like this and so what you can see here is that a curve again it looks like it's kind of squashed together it's got smaller what's actually happened is it has got a lot bigger you can see the negative part of the graph at the bottom twice as big in the lower half it's also twice as big of the upper half it's just that you can't see the top half of it because it's gone off the page now this is more equivalent to enlargement we're kind of stretching and squashing the curve making it bigger and smaller so it is effectively a kind of enlargement but you might be better off calling it a stretch whereas the translation you kind of moved the curve around the next one we're going to look at is y is equal to f of negative x now if the next disabled on the x all this means is that the x-coordinates are going to become negative versions of themselves so if you look at the x intercepts x is three and negative one they're going to become negative versus negative one negative negative one will be positive one and negative three negative negative three will be positive three and so the bottom of our curve is now gonna be on the other side the y-intercept is really important because that's x is 0. negative 0 and 0 are the same thing that's not going to change it we can use symmetry to see on the other side that's going to be that height and then perhaps we look at where it goes off the graph so we're going off the graph at 1 and negative 5. so that's communicable negative one and positive five so a curve it's gonna look like this you'll notice it looks exactly the same it's just on the other side of the graph and effectively this is a reflection in the y-axis so these are reflections and it's a bit odd that you know if it's a y-axis reflection y is the line of symmetry the the mirror line then it's the actual the x-coordinates that become negative y stays the same next one to look at is y is equal to negative f of x and again let me rearrange this uh we can put the negative on the y by multiplying both sides by negative one so what this means is that the y coordinates are going to become negative and logically i guess this is going to be an x axis reflect reflection now again if you look at the coordinates so the y coordinates be negative the x intercepts where y is equal to zero is really important because they're going to stay the same but then the turning point at negative one the negative version of that negative negative one is positive one so the turning point of the curve is going to be the other way around the mind steps at three does that become negative three and then look at the place where it's going off the page at one and negative five they're going to stay the same they're just going to be going at negative 10 instead and so we've got a reflection in the x-axis and reflection in the y-axis to look at now the last thing to look at is something that's interesting it's called the modulus so with the modulus the first one we'll look at is y equals f modulus of x and we use these lines to show the modulus symbol what this means is that the x-coordinates are going to be positive so we have a positive x and y is going to stay the same so let's have a look at our graph where is any x-coordinates positive well the positive on the right hand side which i'm highlighting in purple this part is going to stay the same now we are going to plot some lines in the negative part of the x-axis but rather than being the line you can see the rest of the curve which the negative x-axis values what we're going to do with the negative side is plot the positive x-axis values again and it's actually going to look like this so all we've done all the negative x-values we've put in the y-coordinates of the positive x-values let's look at the version for y that'd be y is equal to the modulus of f of x again the lines showing it's positive so when you look at the coordinates x stays the same and y is going to become positive so we look at the curve we look at the part of the curve where y is positive so i'm going to highlight that in purple and this is all part of our answer it's going to overlap with the previous answer as well now when we get to the bit of the curve where y is negative we just make it be positive so the turning point of the curve at negative 1 it has to be positive y coordinates have to be positive now so it's going to go up to positive one and so it's going to look like this so it's almost like you've kind of bounced off the x-axis when you're plotting the line so there are the four transformations you look at we've got translations we've got enlargement so you know stretches and squeezes we've got reflections and we've got the modulus one final point with these is you might be asked to do a combination of these so it might be that you get y is equal to negative f of 2x and that means you're both going to do a kind of squash on the x-axis by factor two and then you're going to reflect it in the x-axis so again just be careful when you have combinations of these you do both transformations to the x and y coordinates we can look at different things we can do with a straight line and for all these questions we'll use the same pair of coordinates so we'll look at everything we can do with the coordinates negative 9 negative 6 and three seven so what can we do with a line in between those two coordinates first thing we can look at is the midpoint for the midpoint of the line what we need to do is we need to look at the x and the y coordinates and kind of look at words in the middle of those coordinates so the way to do it is for the x coordinates add them together let's see what x1 and x2 so let's say our first set of brackets is x1 and y1 our second set of brackets is x two and y two add the x squared together and divide by two and then do the same thing with the y coordinates that's going to give us negative nine plus three divided by two and it's going to give us negative six plus seven divided by two and so we work that out negative nine plus three is negative six and divide it by two we're going to get negative three then we can do it with the y coordinates negative six plus seven is going to give us one and then one divided by two is going to give us a half so the midpoint the quantity in the middle of the line is negative three and a half next we'll look at the length of the line now to find the length of the line we're going to use pythagoras theorem so we need to know the kind of distance in x and y to make our triangle the distance in x we want the difference between x1 and x2 and then we're going to square that so that'll be the first side of our triangle then we're going to add to that the difference between y1 and y2 and that'll be the second side of our triangle and then we'll square root our answer to find the hypotenuse which will be the length of the line so that's going to be negative 9 take away 3 which will give us negative 12. then we've got our y's so we've got negative 6 take away 7 which will give us negative 13 and what the square root of the answer to that so that's going to give us negative 12 squared is 144. negative 13 squared is gonna be 169. altogether that's 313 and the square root of 313 gives us 17.692 decimal places so that's going to be the length of the line next we can find the gradient of the line the gradient is the change in y over the change in x so we're going to look at our y coordinates it's going to be same as the pythagorean actually so it's going to be the y one minus y two and the x one minus x two and so altogether that's gonna give us negative thirteen over negative twelve and then if we simplify that both negative can make it positive it's going to be thirteen twelfths the gradient change to introduce number 1 and 112. that's how steep the line is all the things we can do with the gradient we can have parallel gradients so the gradient of parallel lines parallel lines are going to have the same gradient so a parallel line will have the gradient 13 over 12 which again is 1 12. perpendicular lines these are lines at a right angle to this line they're going to have a negative reciprocal gradient and to do that you make the gradient negative and you flip the fraction upside down so 13 over 12 will become 12 over 13. and then we can't simplify that any further the last thing i'm going to do is we're going to find the equation of the line and this is the technique you need to find the equation of all kinds of lines given all kinds of clues find the equating the line it's going to be y is equal to mx plus c for the equation of any line so what we want is to sort of shoot into this now we know what the gradient is already it's uh 1 and 1 12 and what we want to do is we're going to put in x and y coordinates from this as well make and choose either the two coordinates i'm going to choose the x ui 2 because it's positive numbers so y is 7 m is the gradient which we had is at 13 over 12 as a fraction and x is three from a second coordinate and so we can work this out to find out what c is so first we can multiply three by three so the seven is equal to thirty-nine over twelve times that fraction by three so we're going to do is take away 37 over 12 away from both sides now if you do that on a calculator what you're gonna get is that c is equal to three and three quarters so we write down the equation of the line it's going to be y is equal to let's write the the gradient as a mixed number this time so 1 and 1 12 x plus 3 and three quarters these are all the different things you could be asked to find from a line note that with this i'll set it with two coordinates but you could start off with a coordinate and you know gradient it could give you a parallel gradient or a perpendicular gradient start off it could be all kinds of different starting points but as you go through these things you're going to be getting the right numbers the equation of a circle tends to look like this so we're going to have an x inside a bracket with a number and that's squared we're going to have a y inside a bracket with a number and that squared has to be equal to a number at the end so what can we take from this well we know that the center of the circle is going to be the coordinate 3 negative 1. so the 3 negative 1 are from the numbers inside the brackets just note the signs have been reversed the radius is the square root of the number at the end so the radius of that circle would be five so that's what you typically see for the equation of a circle now there is an alternate form you can have this equation of a circle looks like this x squared take away 2x plus y squared take away 12y minus 44 is equal to zero and what we want to do really is change this into previous form it's really easy to get the center and radius from the previous form the way we do it is completing the square so we complete the square for x so all we're doing is making this squared into a power one x and then the coefficient on the x term is being reduced by half if we do that though we're not just going to get x squared minus 2x we'll expand it we'll also get a positive one over the expansion from the negative one times negative one so we have to take away the extra positive then we complete the square for the y y squared becomes y 12 y will become six but again negative six times negative six because of 36 we need to take away the extra 36 so now we need to collect up all the number terms so we have a negative one a negative 36 and a negative 44. so the bracket pair looks right collect up all the terms we'll get negative 81 which is equal to zero and we want to want to get that on the right hand side so we're going to add 81 to both sides so when you add up all your numbers reverse the sign and move it to the side that means that now we can figure some things like this so we know that the center is going to equal 1 6. again the sign has been reversed and the radius is going to be the square root of 81 which is 9. so that's how we use the equations of circles one more thing you might be asked is to find the tangent to a circle and a tangent is a line that just touches the circle at one edge it's at a right angle to the radius or diameter so as an example we're going to find the tangent of the equation on a circle and we're going to have a coordinate with it what the tangent at the point 3 6. so 3 6 is on the edge of the circle and the tangent will touch the circle just at that point so how do we do this so the first thing is we want the gradient of the radius the gradient the change in y over x and we actually do have two x coordinates and two y coordinates so let's take the 3 6 first the y coordinate and the x coordinate of 3 6. so where's our other coordinate well we know the center of the circle it's going to be at 5 3. so we can take away 5 3 from this and so this gives a change in y and x between the center of the circle and the point at the edge where the tangent hits so six take away three is three three take away five is negative two so that will give us negative three halves now we can find out the tangent gradient the tangent gradient because it's a right angle it's a perpendicular line it's going to be a negative reciprocal so it would be negative x over y rather than y over x so a green sorry negative double negatively positive and then we flip the fraction upside down so the tangent gradient is going to be two thirds so now we can find its equation we want y is equal to mx plus c we're going to substitute in values for y it's going to be a point at the edge of the circle you can't use the center here so we have to use the three sixths so y is six x is three and now we know the gradient the gradient is two thirds so if we expand the bracket that's gonna give us six over 3 which is 2. so 6 is equal to 2 plus c so take 2 a from both sides c is going to be equal to 4. that means the equation of the tangent is going to be y is equal to 2 3 x plus four let's say you want to do let's just have a little visual of what a tangent looks like someone sketch a rough circle here so the tangent comes by a circle and it touches it at one point and from the center to that point on the edge we have a radius it's going to be a right angle so we have the tangent we have the radius and it's a right angle parametric equation is an alternate way of writing down equations so let's say we have x is equal to t minus one and y is equal to two t squared this is a parametric equation now what does this mean well we're going to go through the process of turning this into the kind of equation you're used to and once we've done that then we can look back and think about why this might be used so what i want to do is want to make both of these equations equal to t so we can make them equal to each other and merge them so for x is pretty simple we can add one to both sides and so we get t is equal to x plus one that's pretty simple now moving on to the second one let's do it a different color so we can color code we have y is equal to two t squared so what we're going to do is we can divide both sides by 2 that will give us half of y is equal to t squared and then we can square root both sides so we're going to get the square root of a half y is equal to t and now these are both equal to t we'll make them equal to each other so x plus one is equal to t this is equal to the square root of a half y this doesn't look like an equation normally the line equations we have y equals so let's try and make y be the subject so we can square both sides that's to give us x minus 1 all squared and a half y on the other side now we can multiply both sides by 2 and that's going to give us y z equals two lots of x plus one squared you might wanna multiply the brackets so if you multiply the brackets you're gonna get y is equal to and remember to multiply by two at the end as well we're gonna get two x squared plus 4x plus 4. now the point of this is with substitution so let's look at substitution let's say we know that y is equal to 1 or we want to find the x coordinate when y is equal to one so how would we do that so with a parametric equation we'll just say well if y is equal to one y is going to equal two t squared but we're saying that y is equal to one so we can solve this we divide both sides by two to get a half and then we can square root it so y is going to be the square root of a half then we can substitute that in for x so we can say x is equal to the square root of a half minus one so very quickly we're getting x and y coordinates a to this even better we have this third variable t and it might sound something like time and if you have values for t then your substitution is going to be even easier now let's have a look at y equals one with our formula at the bottom so we can say one is equal to two x squared plus four x plus four so to get what x is going to be here we're gonna have to maybe rearrange this let's set take one away from both sides to get two x squared plus four x then it'll be minus three equals zero and then you want some kind of method to factorize this you might have to use quadratic formula for example and that's gonna take you a bit of time and then once you do that then you're gonna get your value for x out so hopefully we can see here is a regular kind of equation for a line it's very good to find y coordinates usually exit and get a y coordinate quite easy we're trying to find a x coordinate and substituting y it's not very helpful so again it's just an alternate form to write it so you can do substitutions easier in some circumstances now we're going to look at a harder version of this this was a quadratic let's look at a circle equation as a parametric so we're gonna have x is equal to sine theta plus three and we're gonna have y is equal to cos theta plus seven let's play that minus seven actually so we can see how positive and negatives work now the way we're going to combine these together is our third variable here is theta and we can't really get rid of the sign of the cos to get theta on its own what we can do is use a trigonometric identity sine squared theta plus cos squared theta is equal to one and so we can use this to combine the two terms together by making them make sine theta and cos theta the subjects we'll come back to that in a moment let's rearrange so for x we're going to take 3 away from both sides that will leave leave us with sine theta is equal to x minus 3. well the cos gonna add 7 to both sides so that'll be cos theta is equal to y plus seven so then if we bring back our trigonometric identity we want sine squared theta to substitute it now we have sine theta sine theta is x minus three so we want the sine squared theta we just need to square it then with squash squared theta we have cos theta it's y plus seven so all we need to do is square it and now we combine the two parametric equations together into a single form and you might notice that's actually the equation of a circle now okay the bit of rearranging but we've got it straight away so now we can say things about the circle we can see that the center of the circle is at three negative seven and we can say that the radius of this circle is the square root of one which is one and again there's different benefits to both forms you see combined together it's really easy to find the center in the radius whereas when the repair is gonna be easier to substitute in values for theta or even values for x and y and rearrange but rearranging with a full equation is going to be much more of a mess before we look at binomial expansion we have to look at pascal's triangle now the triangle has an infinite number of rows but we are going to have a look at the first eight rows now it starts off with a one and for the next row we're gonna have one and one now it becomes clear what's happening with the next film we have one two and one and what's happening with this triangle is that each number is the two numbers above it added together so the two is the two ones above it added together on the edge you only have access to a one and a blank space so they stay as one so for our next row the edges of pascal's triangle are going to be ones but then we have one plus two is three and two plus one is three and so we can generate the triangle like this each time adding up the two numbers above so one plus three is four three plus three is six and three plus one is four so let's continue so we have each times and the numbers above and you can see an alternate rows you have uh two identical middle numbers so it's an even numbered kind of number of terms that you have then the middle two numbers will be the same if it's an odd number like this one then the middle two terms will be different so let's just generate the last few rows you'll also see that this is symmetrical so the numbers on the left are the same as the numbers on the right now what it's used for is to find the patterns when you expand brackets so let's say we had a quadratic we had x plus one squared if you expand this quadratic you get x squared plus two x plus one and compare that answer to the second row of pascal's triangle we've got a one a two and a one in the triangle and then you look at the coefficients here we've got a one x squared a two x and then a one so it's matching that pattern so then when we look at x plus one cubed that should give us x cubed plus three x squared plus three x plus one again using that pattern and then we can do well it's the power of four we can use this pattern we're gonna have x to the power four plus four x cubed plus six x squared plus four x plus one now as well as the coefficients on the x matching the rows of pascal's triangle so for example the fourth row we've got one four six four one in an expansion we've got one four six four one also have a look at the powers now the powers on x are going down each time so we've got power 4 power 3 power 2 power 1 and then we don't have a pair at the end which means it's going to be a power 0. we can use this pattern to write a formula so if you've got a plus b to a power so like that's what we've just done a b x b will be one then we've got the various powers the formula is going to be n over k multiplied by a to the power of n minus k multiplied by b to the power of k and what this does is this will generate the terms in the expansion so let's say in the kind of power four expansion we just did let's say we only wanted the term on x squared what would we do so what we would do looking at this formula is first we have n over k now what that is going to give us n is the power so this is a powerful expansion and k is going to be the pearl looking at so we're looking for power 2. then the a that's going to be our first term to the x so that's going to be x to the power of n minus k so 4 minus 2. and then b which is a 1 here that's going to be power of k it's going to be the power of 2. we do this multiplication we should get six x squared now the key thing here is what is this bracket about we've got kind of looks like a vector we've got four over two now what that does is that generates the terms in pascal's triangle so if you've got n over k this uses a formula where it takes the factorial of n that's m multiplied by all the numbers below it all the whole numbers below it now it's divided by k multiplied by n minus k and it generates the part of the triangle you're looking for now the n is the row and k is the column and remember columns always start with zero so we label this for the bottom row we're gonna have column zero one two three four five six seven and eight but obviously that's gonna be different for each row again each row starts with zero then you just count your way along so when we're saying four two we say you want the number in row four and column two so look for row and four then we look for column two so zero one and two and we can see that's going to give us the six so the the kind of vector here the row and the column that's going to give us 6. x to the power of 4 minus 2 is x to the power of 2. then we multiply it by 1 squared and 1 squared is 1 so 6 times 1 is going to give us 6. and so you can see by applying the formula we've got the 6x squared straight away so it's useful for finding particular terms you might be asking examples to find the last two terms on the first two terms for example or perhaps a term on a particular power or like we just did here last thing to think about is that you don't need to draw a triangle out for every single question so we're talking about the kind of vector notation with the n and the k is actually a button for that on your calculator and the button for it is the ncr button so you type in n there's then a button with n c air on it and then you press k when you type that into calculator you should see n you should see a c and then you should see the k and again that's the method for your calculator to know which row and column of pascal's triangle you want arithmetic progressions are some of those kind of basic sequences you'd find on gcse but you might have slightly different terminology you might say u n which just means the nth term so let's say the nth term is a half and plus a third so you'd have to be able to generate the sequence so to generate the sequence uh n would start off with one the first term so you'd have one half plus a third and you could without using your fraction skills but a half plus a third is five sixths then for each pair of sequence the way it's going to scale up is it's going to go up by one half each time so for the second part of the sequence n will be two that'll be two halves which is the whole one plus a third that'll be one and one third the next part sequence n would be three so you have three halves plus a third and what you might find here is rather than working mixed numbers it might be better to work in improper fractions now a half is going to be the same as three sixths so just add on three sixths each time so eight six is the same as one and a third but it's just easier now to be adding on the three to the numerator each time to generate a sequence and then the sequence would carry on forever you might be asked for specific parts of the sequence so you might be asked for say u10 for the tenth pattern sequence that's gonna be a half times ten plus one-third now if you did this with mixed numbers a half times ten would be five and five plus a third is going to be five and one-third if we can keep it consistent what we're looking at before then we could say well this is 16 thirds which would be uh 32 6. so be flexible in the way you can write your fractions depending on the context because if you have the denominators all the same for all parts of the question it's easier to compare so you see that 32 over 6 is a lot larger than 17 over six the fifth term now i'm looking at five and one third and i can't have to guess how it relates to the other parts if you are asked to generate the nth term from the sequence then look at this sequence you should be able to see that it's going up by 3 6 each time which is the same as another half each time and so if it does the same thing each time and it does then you know it's going to be a half n so the number on the end the coefficient on the n is always where it's increasing by each time now if that was true it was a half n the first term would be a half times one which is a half and it's not it's five six so you think what about onto a half to get the first term five six now five six and a half the denominator is different what half is three six heading from three six to five six we've gotta add on two six which is the same as a third so that's where the third is coming from in this so this is kind of recap how it works at ggs these are more complicated uh fractions involved which is quite common now another thing you might be asked to do is to add up a certain number of terms in the sequence and we've got a formula for this so we have this symbol this is the sigma symbol and the sigma symbol just means sum and it's going to add up all the things we'll look at here now on top of this symbol we have n this is a number of terms we're going to add up and then below the symbol we've got a number and the number underneath that's where the uh the adding up starts very often it's going to be equal to one next to this you have the nth term so we've got the previous question we could have a half n plus a third and this is the kind of terminology you'll see in the exam and it's asking for the sum the formula for this is s is equal to n over t multiplied by 2a plus n minus 1 multiplied by d and so we have to think about where all these different letters come from so n is a number of terms to add the a is the first term of sequence so we know from our previous work here the first term is 5 6 well that's something you can generate from the nth term by saying that m is equal to one d is what we call the common difference the common difference is what's changing each time so in the sequence it's going up by a half each time so the common difference is a half so let's say we were given a question on this let's say we said right we're going to sum up we want the first five terms the one that i just rotate here i'm going to start off adding from the first term but in the brackets you actually have the nth term so we would substitute into the formula s is equal to the first five terms divided by two two lots of the first term a so the first term is five six two lots of five six plus n minus one so five minus one multiplied by the common difference d which is a half so we can start to work this out so five divided by two is going to give us uh two and a half two multiplied by five sixths to be ten sixths i'm gonna add on to that five take away one is four so four multiplied by a half we can continue working through this so we have two and a half multiplied by ten sixths plus now half of four is two so if you do uh ten six plus two and then multiply it by two and a half the answer that i'm getting is six point one six recurring which we can write as a fraction one place is occurring is a six so it'll be nine and one six so if you want to spend some time on this you can continue working out with fractions or you can do what i just did there and just work it on the calculator as a decimal and convert it back into a fraction at the end you're probably better off keeping it in fractions the entire way with your working but i'm just going to speed this up because it's just a revision slide so now what i've done here is i've added up the first five terms now let's have these written outs let's see this actually works so 5 6 plus 8 6 plus 11 6 plus 14 6 plus 17 6 should give us 9.16 occurring and on my calculator it does so we can see that the formula should work one last detail is that for this formula we have k equals one the starting point of where we are adding from is the first term now if the starting point isn't one let's say it's two then you have to take the first term off your answer because this formula doesn't take the first term the kind of starting point in two accounts if you had to start counting at the third term they would do your formula that'll be answers for all the terms up to where you're looking for and and then if you want to start counting the third term you have to take away the first and second terms so just be aware of your starting point if you need to take anything away if k was quite large let's say k was equal to five you could actually use the formula to find the sum of the first four terms and use any as four and then you would take that away from your answer which has all the terms added together up to wherever your endpoint was geometric regressions again you might see some notation for the nth term which you might have un is equal to and let's say for example you have 2 times 3 to the power of n so this is going to describe a sequence in the first term of sequence n will be 1 so that would be 2 times 3 to the power of 1. well per 1 doesn't change a number so 2 times 3 is 6. when n is 2 this would be two times three to the power two which is two times nine which will give us eighteen then we'd have two times three to the power of three which is fifty-four then we'd have to the power of four 162 and the power of five which is 486. these numbers are starting to get bigger quite quickly as usual for sequence the number of numbers and sequence could go on infinitely i'll actually be given the limit for the sequence now what's happening in the sequence each time we are multiplying by three this is a times three sequence so geometric progressions are about sequences where you multiply or devise arithmetic for sequences are sequences where you add or subtract each time and each one uses different formulas now for this one if you can see this times by 3 each time you write 3n that's where that comes from they think well if for the first term 3n be 3 to the power of 1 which is 3 how do i get to the six and so that's where we have to then add on the multiplying by two to get to our first term you might be asked for specific terms let's say you ten will be the tenth term so you do two times three to the power of ten so we would take the calculator three to the power 10 multiplied by 2 is going to give 118 000 and 98 so again when you're doing these kind of sequences they get very very large very quickly now again it's a special notation to look at so we've got the sigma symbol for the sum and on top tells us how many parts these ones want to add together and the k underneath will tell us the starting point commonly you'll see k being equal to one after this you'd have the nth term so you would have whatever you're multiplying by you know the thing you're multiplying and you would have the power the formula for the sum is the sum of the first n terms is a multiplied by 1 minus r to the power of n divided by one minus uh we need to know all the different letters mean so m is the number of terms to add a is the first term in this case it's six for our previous example uh is what we call the common ratio this is basically doing the sequence each time so in our last example we're multiplying by three each time so the common difference will be three it's the same as the number in the x term that's to the power let's look at what got the k is just the starting point of where you're at so we're going to start the first term usually and if k equals 1. we'll talk about other numbers later so let's try adding up the first five terms of our original example now we can actually do it because we've got the first five written out but what we're going to do is just test if the formula works so the sum of the first five terms the first term is six we're multiplying by one minus the common difference three to the power of the number of terms which is five so let's work at the numerator and denominators so three to the power of five is going to give us 243 so one take away that will give us a negative 242. then one minus that's one minus three that's going to give us negative two so we work this out negative 242 divided by negative 2 gives a positive 121. then we multiply that by 6 for our answer and we get 726. now let's go back to our last example let's add an open see what we get the 6 plus 18 plus 54 plus 162 plus 486 all that together it does give us 726 so we can see that the formula works another important example of these is when the nth term is going to give us some kind of fraction so we could have something like 3 times 3 to the negative n an alternative way of writing that is going to be 3 times 1 over 3 to the power of n now if we generate this sequence it's going to give us the first term energy one so three times one over three that's gonna give us one then when n is two three to the power two is nine so it's gonna be three times one over nine which is three over nine which is a third then when n is 3 3 to the power 3 is going to be 27 that'll give us 3 over 27 which is going to give us 1 over 9. what you might notice the denominator is getting 3 times bigger each time so what's happening with this sequence is each time we're actually dividing by three and another way of doing divide by three is to write it as a fraction which is why the denominator is actually getting three times bigger each time now with these you have another kind of formula and this formula is sum to infinity because what's going to happen is you know it's with the sequence one third and negative one over 27. they're getting smaller and smaller and smaller these numbers so that's the limit to how big this sequence can get even if you add up all of these numbers until infinity because the numbers will start getting infinitely small so this formula is a multiplied by 1 over 1 minus r so we're lacking the n part of this sequence so if we try and say with this example the sum to infinity a is still the first term which is one we're multiplying by one over one minus r which is the common difference now because it's dividing by three each time we write it as a fraction so we're going to write it as one-third so now we can start to work this out we have one multiplied by the bracket and that's gonna be one over one minus a third is two-thirds now if you get a triple-decker bracket like this this is really bad we don't want to do this now you can test it out on a cartilage and see what happens but what you can basically do in this situation is put the fraction upside down you get three over two so the card click let's say types in one divided by two thirds let me write it as a decimal 0.6 occurring it's going to give you 1.5 as an answer and 3 over 2 is also 1.5 so what the solution infinity tells us is that with the sequence 3 times 3 to negative n no matter how many items in that sequence you add together you will never get higher than one and a half because each time the numbers get smaller and smaller and smaller and smaller and smaller the last thing to mention is that for all these questions our starting point k equals one if the set point k was something else let's say it was two or three or four they would use these formulas and we would subtract the terms that we don't want so let's say that k was equal to two once it started adding from the second term in the sequence you would follow this as normal then you would take away the first number the first part of the sequence radians are an alternate way of measuring circles so usually with degrees a circle is 360 degrees now when we look at radians a circle is 2 pi radians the reason for this is that 360 was just a really good number people chose because it's got lots of factors you can divide lots of different ways so 2 pi is a way to kind of relate the measurement of a circle to some actual mathematics there's actually a reason why we're using pi whereas 360 was just kind of you know used because it was a good enough number now you could convert between the two so to get from 360 to 2pi you need 2 divided by 180 360 divided by 180 is 2 and you need to multiply by pi to get the pi on the end and that will convert any number in degrees to radians you can also go the opposite direction and convert from radians to degrees to get from a 2 to a 360 you need to multiply by 180 and then we'll need to get rid of the pi so we divide by pi so do those two things in either order you can convert anything in radians into degrees now as well as being able to convert them perhaps using the calculator for this there's also exact values you should know so the exact values are 0 degrees 30 degrees 45 degrees 60 degrees and 90 degrees now if you convert these into radians you're gonna get zero radians pi over six radians pi over four radians pi over three radians and pi over two radians and the reason why we've got pi in there are not a decimal so for example pi divided by 2 is 1.570796327 that number will continue more decimal places it's a rational number it's not going to end there's not going to be a pattern so rather than writing out the whole thing or as far as we can get or even rounding it to be shorter if we write the pi in there it means we don't need to round so we have our degrees and radians now as well as memorizing that we need to memorize what the sin cause and tan these are sine theta cos theta and tan theta so at zero sine is zero cos is one and tail is zero at 30 sine is a half cos is root three divided by two and tan is one divided by root three at 45 degrees sine is one divided by root two cos is one divided by root two and tan is one at six degrees sine is root three over two cos is a half and tan is root three they're 90 degrees sine is one cos is zero and tan is not defined there's not a solution for it i guess in a way you could say it's infinity you look at sign and calls you'll see silent chords have got the same values but in the opposite order so if you can remember the sign values if you flip them over you've got the cost values knowing these exact answers is useful for a few reasons so it can't be useful if you're working with certs or it might be useful if you're working without a calculator and so you want to work with a full level of accuracy without rounding trigonometry in a right angle triangle is going to involve a second angle being labeled then we've got the hypotenuse the longest side the adjacent is next to both labeled angles and the opposite which is the the one opposite the uh the labeled angle that isn't the right angle now if you have some values for these then you're going to be able to use trigonometry to find the other missing value also should remember that the angle we can label that theta we have the acronym so car tower that helps lex method and each one of these references a triangle so we have the solar formula triangle and from left to right you just go s o h we've got the car formula triangle and again you just go from left to right c a h and the toe for the triangle left to right t o a now the o the h and the a correspond to which ones of the o h and here you have in the original diagram the s c and t refer to the angle if you want to find a missing angle theta then look at the former triangles you cover up the s c or t you look at what's left over and you should have two letters on top of each other so for example theta could be the inverse sine of o over h that's what you get for the first formula triangle or it could be the inverse cos of a over h or the angle could be the inverse tan of o over a that's what it means when you've got the kind of the letters on top of each other if you're looking for the opposite then we look at the form of triangles you cover up the opposite and whatever's left over is the math you do so from the first form of triangle the opposite if you've got s and h next to each other that means the sine of the angle the sine of the angle theta multiplied by the hypotenuse and you've also got that in the last form of the triangle you cover up the opposite what you've got left over is the t for tan so you've got tan of the angle multiplied by the adjacent for the side a you cover up the a in the form of triangle so the first one if you cover up the a you're going to get that the adjacent is equal to the cosine of theta multiplied by h because next to each other but in the third form of triangle is the a is a different position and it's all over t see the adjacent will be equal to the opposite divided by the time of the angle last one to think about is the hypotenuse so again you cover up the hypotenuse to find it and you look at what's left over in the formula triangle so we've got h in the first form of triangle you've got o over s so that means that the hypotenuse is going to be equal to the opposite divided by the sine of the angle you've also got h in the second form of triangle if you cover up you've got a over c so it's going to be the adjacent divided by the cosine of the angle now you're only going to use one of the formal triangles and it depends on what you've been given what's been labeled so it might be you're not given the hypotenuse at all you're only given the opposite the adjacent so you can only use the third form of triangle in that example and so you'd only be able to find out the opposite of the adjacent or the angle depending on which of the two of the three you're given as a clue now you can also use trigonometry in non right angle triangles and we have some formulas for those now the way we label a non-right angle triangle is that we would give a letter for each one of the angles so we could call them a b and c they're another letter for the sides what we do is we make the size a lowercase of the opposite angle so we've got side a side b and side c so lowercase for a side and a capital letter for an angle so that brings us to the sine rule the sine of an angle divided by the opposite side is equal to the sine of any of the angles divided by the opposite side so these are all in the same ratio and we can use this to find missing sides and angles you can also do this the other way around so you could have the side divided by the sine of the opposite angle another rule related to this is the cosine rule the cosine rule is related to pythagoras theorem so in pythagoras theorem c squared is equal to a squared plus b squared but this only works for a right angle triangle now if you don't have a right angle triangle you need to make an adjustment for the angle being different and that adjustment is taken away 2 multiplied by a multiplied by b multiplied by the cosine of the angle c there's also independent of the names of the variables so we could you know need letters in another way around and so we can write it different ways we can write a version that's b squared equals and a squared equals so the way it works is the thing you're looking for it's going to be the two other let's add together so you're looking for b it's going to be a squared plus c squared you're looking for a it's going to be b squared plus c squared then we make the adjustment it take away two of the squared signs so two a c or two b c and then the cos is always the angle related to the thing you're looking for so you look at the side b it'll be cause angle b you're looking for side a it will be cos angle a the last thing you need to know that for this is how to find the area of a triangle using trigonometry now like the cosine rule like the pythagoras theorem this one's related to the area of the triangle the area of a triangle is equal to a half of the base times height now we don't have a base and a height in a non-right angle triangle because the base and the height have to be at right angles to each other so let's just call an a in b so that's the formula for a normal triangle within the right angle if it doesn't have a right angle you have to make an adjustment the adjustments really easy you multiply it by the sine of the angle of the side you're not using so are you using a and b as their sides what the sine of c and again just like the cosine rule you can do this with the other letters so a could be half of ac sine b or you could have half of b c sine a plus it's two sides and then the unrelated angle so here you've got all the kind of g to c trigonometry formulas on one page and also don't forget pythagoras theorem you'll need to use that sometimes c squared is equal to a squared plus b squared trigonometry equations combining equation solving skills with trigonometry now you might have something like sine theta is equal to 0.4 this is something you can solve using equation solving skills now we want to get a theta on its own for this question so the first thing to do is that we can do the inverse of sine to both sides when you usually kind of form a triangle trigonometry you sign to minus one or cos to the minus one this is practically what you're doing so if we do the inverse sine of 0.4 we get 23.6 to one decimal place and you might think great that's the answer we do trigonometry with triangles this is as far as you go however there are actually multiple answers for this now the way you're finding multiple answers is you need to sketch a sine curve so the sine curve is going to go from negative one to zero to one we need to remember the way our understand curve goes now what you could do to calculate press sine zero and that'll give you a coordinating point sine zero zero then you could do say well what's sine and ninety the sine of ninety is one then you think well what sine of we got up in nineties what sine of 180 sine of 180 is zero up another 90 sine of 270 is negative one and sine of 360 is zero so it gives you an idea of where the sine curve goes then we can join it over the curve and here we have a rough sketch of a sine curve now we might want to label the uh the x-axis i'll get the y-axis labeled which might think we've got zero degrees 90 180 270 and 360. now our answer is at 23.6 degrees and that corresponds to 0.4 so 0.4 to be just below half and what we can see is in between 0 and 360 0.4 cuts through the curve in two different places that means there's actually two answers the first answer we have is right at start so it's in between 0 and 90. so this is our 23.6 we can think well from the start we've gone in 23.6 degrees now the symmetry with these curves if we're looking between 0 degrees and 180 degrees there is a line of symmetry through this curve so over 23.6 on the left hand side using symmetry the second value is also going to be 23.6 in but this time going from the right hand side which is at 180 so 180 and take away 23.6 to go 23.6 before it will give us the second answer which is 156.4 degrees so this is all about yeah using your solvent equation skills but also recognizing that we do inverse sin cause our tan there's going to be multiple answers so let's look at the version with cos so let's look at cos theta of let's do negative 0.7 so we would do the inverse of cos to both sides and that would give us that theta is equal to so we'll do the inverse cos negative 0.7 and i'll give that first answer 134.4 degrees now again there is going to be a second answer between 0 and 360. and we're going to have to draw something to find it so again you have to do this yourself you have to sketch the curve between negative one zero and positive one then with angles it's good to go with increments of 90 degrees eventually at first stage we remember how the curve goes if you don't remember the curve use a calculator type in cos zero which will give you one type in cos 90 which will give you zero cos 180 will give negative one and just putting these in will give you an idea of where the curve goes and then you can sketch it so it's going to go like this now cause 0.7 it's gonna be down at the bottom clustered negative one and zero and we can see again it's gonna cut through the curve in two different places now we have the answer between ninety one eighty this is a hundred and thirty four point four think about how we did last time we've gone in from the left hand side 134.4 degrees now the line of symmetry in a cos curve is different in between 0 degrees and 360 degrees the line of symmetry goes through the middle at 180. so we say well 134.4 into the left-hand side let's do the same thing from the right-hand side this time i'm going to take away from 360. and we take 134.4 away from 360. we've got our second answer which is 225.6 degrees we can do this with tan as well so let's have a look at the tan of 0.5 so we do the inverse of 10 to both sides and that gives us that theta is equal to the inverse times 0.5 which is 26.6 degrees again this is only our first answer so we need to draw the graph to figure out what the second answer is the calculator only gives one answer she'll set up a y-axis v curve between negative one zero and one set up an x-axis zero degrees ninety degrees one eighty going up in 90 degree increments and then sketch the curve and again if you're not sure how to sketch it you can type tan 0 into calculator which will give 0. so first coordinate 1090 will give us a math error that means there's an asymptote here the line does not go through 90 degrees then we can do time 180 which is zero degrees again turn 270 again give us a math error and tan 360 is also going to give us zero now this hasn't really helped us draw a curve has it she might look at in between values let's try going up in 45 degree increments tan 45 will give us one so we know the curve is going to go up like this add another 45 we get 90 and another 45 to 90 and we're gonna get 135 so tan one three five is going to give us a negative one then in between 180 and 270 and 45 get 225 the tan of 225 is a positive one and this gives an idea of where the curve goes so it's going to be something like this now i want 10 to be 0.5 so to go through at 0.5 we can see that we'll cut the curve in two separate places now we have the answer right at the start which is 26.6 there's a special thing with a tank curve and it's that it repeats every 180 degrees so all you need to do is take the 26.6 add 180 to it and you're gonna get another answer and that's 206.6 degrees now the next thing we need to talk about is a range all of these questions have a range that's in between 0 degrees and 360 degrees now you won't always be given this sometimes you might have between 0 degrees and 720 and so for 720 you would make these curves go twice as long so you draw them going up to 720 you could have it between negative 180 and positive 180. so you draw the bit of drawn here then you'd also go into the negative section as well now regardless of how big this is if you look between 0 and 360 you can accept because these curves they always repeat as well every 360 degrees they repeat and that's all three of these repeat every 360 degrees so with the two answers that you've got as long as you add and take away 360 degrees from them you're going to get other answers because the big thing with these is that they kind of go on forever so since the curve's gone forever we need to limit it somehow again with zero 360 the most common but you might have others like i mentioned the other thing is all of these are in degrees you might also be asked this question in radians there's two ways to do it in radians you can do what i've done here and then we get your final answers you can multiply by pi and divide by 180 and now your answers are in radians the other way is that you can use the mode button on your calculator and set your calculator radians mode you'll be at a little r on the calculate display it'll give you the answers in radians straight away just make sure once you've done with that change it back to degrees mode a little d display otherwise you get confused next time you use trigonometry and you sound content by giving you answers are incorrect you don't expect because it's in the wrong mode with trigonometric identities we can change some kind of expressions trigonometry into others so for example if you take sine and you divide it by cos you will get tan so you can convert tan into silent cards or vice versa using this method then we've got more complicated trigonometry words so we've got cosec cosec is equal to one over sine then there is sec sec is equal to one over cos and there is cot which is equal to one over tan and these work the other way around as well so sine is equal to one over cosec cos is equal to one over sec and tan is equal to one over cut now with the tan is equal to one over cut we can write another identity which is the following cut is equal to one over tan therefore is equal to cos over sine and the whole thing with this when you have one over something what you've done is you've taken the reciprocal of the number and so we take the reciprocal of kind of like a normal number number on its own then you're gonna get a one over but and it kind of turns into a fraction but it's already a fraction like sine of cos then we take the reciprocal it just flips a fraction upside down this is our first set of identities the next step is that sine squared theta plus cos squared theta is equal to one and by rearranging this we can find out a few things so for example sine squared theta is equal to one minus cos squared theta cos squared theta is equal to one minus sine squared theta then we can get versions with set cosec and cot so tan squared theta is equal to sec squared theta minus one then we can say things like sec squared theta is equal to tan squared theta plus one quad squared theta is equal to cosec squared theta minus one and cosec squared theta is equal to cos squared theta plus one so this is another method where you can take same cause of time and you can kind of convert them into each other depending on what you need we also have identities around about adding angles together so sine we've got angle a plus angle b is equal to the sine of a multiplied by the cos of b plus the cos of a multiplied by the sine of b and that works if you take away as well there's a version for cos because of a plus b is equal to because of a multiplied by the cos of b take away if it's the opposite sign the sine of a multiplied by the sine of b and then if you're taking away the angle that'd be adding the angles instead then there's a version of the tan if you have tan angle a plus angle b that's going to be equal to the tan of a plus the tan at the b and then it's all going to be divided by one take away so the opposite sign the tan of a multiplied by the tan b then you take away the angles instead it'll be take away the numerator but it'll be an ad for the denominator then we've got the double angle formulas so the sine of 2 theta is equal to 2 sine theta multiplied by cos theta it's also equal to so we've got another alternative two lots of tan theta all divided by one plus tan squared theta then we've got options for this for cos and tan so cos of two theta is equal to cos squared theta take away sine squared theta it's also equal to two lots of cos squared theta take away one and one minus sine squared theta and it's also equal to one minus tan theta all divided by one plus tan theta so we have lots of different options for changing the forms of these there's also one to ten let's pop that up at the top we've got a bit of room tens of two theta is equal to two tan theta divided by one minus tan squared theta so we've got a big list here of different genomic identities some of which you're going to have to be able to use to replace some trigonometric terms with others better things to do here is look at the list to write it down try and find the formula booklet for the exam you're doing and then make sure which ones of these you need to remember for the exam and which ones are going to be given to you in the form of booklet what you'll find is though you don't need to remember the whole list because some of these rearrange to make other ones so you can remember a smaller list of these to make the rest with differentiation from first principles you might have a curve that's something like y equals x but it'd be actually useful to look at what that graph is going to look like so i'm going to draw some axes y equals x squared is going to look something like this and what we mean by differentiation if you want to know what the gradient of this line is now the gradients how steep the line is and you'll notice that you know at the bottom it's going to be less steep so that's highlighted in green and at the edges it's going to be more steep so let's highlight that in pink so the steepness of the line depends on where it is so how can we give it a gradient how can we say how steep it is so what we can do is we can find the gradient of just one pair of the graph now it's not enough to do a single coordinate we're going to look at the gradient in between two different coordinates it's not exactly a lineup it's gonna be a straight line it's not gonna be a curve will give you a rough approximation so what are these coordinates going to be well it's called y equals x squared so if you look at where the x coordinate is of the first one you call that x puts on the x axis then the coordinate on the y axis that corresponds to it that's going to be x squared it could be y but we know that y is equal to x squared so our first coordinate is going to be x x squared so we look at the next coordinate now if the next coordinate is going to come down a little bit further along the x-axis so what we do is we can call this x plus h we've got a distance of h in between x and an x-coordinate on the x-axis and we'll just say it's an unknown number because i don't really know how far along that we've gone so the x coordinate is going to be x plus h what's the y coordinate going to be so again y is equal to x squared so it's x plus h we need to square that so we take our x plus h and we square it so let's write it in the coordinates and having the brackets and squaring the entire things really important because then that gives us a quadratic that we can expand later to run so now we have our two coordinates we're going to think about how we find the gradient now in general to find the gradient of anything it's the change in y of the change in x the way that works with two coordinates is you take one of the y coordinates and you take it away from the other y coordinate it doesn't matter which way around the two coordinates are as long as consistent so whichever you pick first for y also has to be the first pick for x and you do the same thing for x so what we can do is we can substitute into this formula to find out the gradient in between these two coordinates so our second y coordinate was one in brackets x plus h whole squared and then we're taking away the first y coordinate which was x squared we're dividing that by the second x coordinate which is x plus h and then we're taking away from that the first x coordinate and so what i want to do now is expand the brackets and clean this up so you should know how to expand a quadratic x times x is x squared x times h is going to give us x h h times x is also x h we have two of those and h times h is h squared and then we're taking away x squared we've already got an x squared so that'll cancel out to give us zero then for the denominator x take away x to get a 0 so we're only going to have the h left we can also divide through by h because both terms and numerator i've got a h and there's a h on the denominator so we take a h out of both terms 2 x h becomes 2x and h squared becomes h and so we've got an expression that represents the gradient in between the two coordinates now what we're going to do next is a little bit tricky so if you look at the graph the distance to the two coordinates on the x-axis was h and that was just a random number chosen we don't even know what the number is that's why we labeled it with a letter now imagine you make the distance between x and x plus h smaller you make it half as big is that going to change our calculation here and it's not it just means that h is going to be plus a smaller value what we can do is we can reduce the distance between the two quad bits more and more and more and the more we reduce the distance between them the more accurate our gradients going to be because at the moment it's a straight line on a curve but the closer together they get it's going to be you know less inaccuracy making that straight live on the curve and what we do is we say that there's going to be a limit as h approaches zero and so what's going to happen is eventually these two quads can move them so close together that the value of h is completely irrelevant and if we have h approach 0 then we get only 2x on its own we get rid of the h and what you'll find is that having h approaches 0 go away what's left over the 2x actually represents the gradient of the curve now it's not a number like you'd have before with a straight line you know the gradient's going to be a whole number it's going to be a two steepness of two a five statements of five zepa or you have steepness over quarter and that wouldn't be very steep at all you get a number this has got an x in it and what it means is if you substitute in the x then it's leave the gradient no matter where you are on the curve so an example at the bottom where it's not very steep x is going to take a smaller number so i'm just going to highlight that in green like i did before so let's say hey let's say x is 1 then 2 times 1 is 2 it's a green of 2 which is you know vaguely it's not too steep if you're right in the middle the flattest part that's when x is zero and any gradient zero that's not very steep at all now let's look at the edges if we imagine these edges are when x is like 10 or negative 10 we substitute 10 in for x you get 2 times 10 is 20. well that's really really steep and so you can see it working you substitute in x you get the gradient at that point so this is how you differentiate from first principles now we've done this for y equals x squared and that's why our second coordinates were squared now you can do this for other things as well you can even do it for trigonometry you know it's more complicated curves you for cubics you know y equals x cubed plus two x plus eight whatever it is and all that happens is the x coordinates be exactly the same x and x plus h but the y coordinates will have a different transformation based on what you're doing but if you just follow the process at the end as long as h approaches zero then you are going to get to the gradient now with differentiation from first principles y equals x squared is differentiated to two x and i said when you differentiate you get the gradient of the line so the gradient of y equals 2x is the difference in y over the difference in x and this is where we get our differentiation notation from difference in y over difference in x now you don't actually have to go through differentiation from first principles to every single question there is a shortcut and if you look at the question the answer you might be able to see it you might be able to see that the power on the x has gone down by one we've gone from x squared to x to the power of one there's no one shown but there's no power on an x it just means the power of one so that's hidden information the other thing is it was a power two and now we've got a two at the front and there is a connection there so let's look at the general rule for differentiation so if we say that y is equal to a x b so the power is b a is a coefficient on the x then the difference in y or the difference in x the gradient is going to be equal to so you multiply the a and the b together so we can write a b for that and then the x the power is going to go down by one so it's the power b minus one so we if we apply that to what we've just done we can see the power goes down by one now with y equals x squared there wasn't an a there wasn't a coefficient on the x or was that if there isn't an a it's just gonna be one we have one x squared so you do have to deal with hidden information in these questions we've got the hidden one in the one x squared and we've got the hidden power one in two 2x so again two things the power has gone down by one and we've multiplied the coefficient on the x by the old power you can also differentiate a second time we call it d2y over dx squared so if y equals is a line and differentiate it once dy by dx gives you the gradient d2y by dx square the second derivative that's the rate of change of the gradient is how the gradient changes all the time you just follow the same process so we're going to reduce the power by one so x has gone down from two to one now it'll go down to zero and then we multiply the coefficient at the front the two by the old power one and two times one is two so it's actually going to stay as two now when you have power zero anything to the power of zero is one so effectively this is saying two times one so all together it's actually just two now with this we've only differentiated one term we can differentiate multiple terms together as long as i separate by plus and minus so we could have y equals x cubed plus two x squared plus three x plus four just as a random example so we differentiate once to get the gradient here's what we do each time the power on the x is gonna go down by one so x to the power three becomes x to the power of two x to the power of two becomes x to the power of one x to the power one becomes x to the power zero and we'll look at the last one in a moment but i guess you could stay with the plus 4 it's going to become x to the power of negative 1 because there's no x on 4 so it'll be x to the power of 0. so that's the first bit tell with the second bit is you multiply the coefficient on the x by the power if there's no coefficient it's going to be a 1. so one times three is three two times two is four three times one is three and then with this one this was x to the power of zero so four times zero will actually give us zero now this isn't really how we write algebra so there's some information gonna be hidden so just like it's a one x cubed and it's a three x to the power of one the hidden information here is we're not going to show the power one on four x to power one we don't show x to the power of zero because it's just equal to one and three times one is three and we're also not going to show anything with a coefficient of zero because it's got a coefficient of zero he times him by a zero so no matter what the rest of it is it's going to be zero so we don't write it down so now you might be able to see the final answer here it's going to be three x squared plus four x plus three now it's worth just looking at this at the moment and just looking at kind of which terms disappear so for example the plus four if you've got a whole number on the end of something and you differentiate a whole number you get zero because whole numbers have got x to the power of zero on them effectively even though we don't write it down so you're multiplying them by zero so they go away the next thing is if you differentiate like three x so a number with just an x to the power of one then you just drop off the x three x has become three it's only squared and upwards where you can use the full method and show all the information we multiply the coefficient by the power and you can reduce the power by one just take note that when you differentiate some of the squared it's going to become power one and we don't write down power one so you just write x so there's a few things to look out for there now you wouldn't write all the hip information down every single time so if we differentiate the second type for the rate of change of the gradient you're gonna be able to do this in your head without any extra working in really so we're going to do 3 times 2 is 6 reduce the power by 1. 4x is going to become 4 and the 3 is going to go away and it's actually that quick you don't need to write all the rest of it down so again whole numbers disappear it's gone anything with an x the x is going to drop off and then with pose two or higher you go through the four method now there are some special cases in differentiation because it's not just kind of normal algebra you're gonna be differentiating there's also another way to show it we're rather than writing y equals you can write f of x function of x is equal to then the differentiated version would be f dash x so you can see it written both ways so firstly you can differentiate trigonometry so if you have sine of x and you differentiate it you get cos of x you can also change the angles the x is the angle now let's say we've got a more complicated angle let's say look at the sine of 2x we differentiate the angle is going to stay the same but what is going to happen is the coefficient on the x is going to pop out onto the outside as well so you have sine of 2x differentiate it you get 2 cos 2 x the same thing happens with cos so let's say we've got cos of 3x when we differentiate it when you differentiate cos you get sine the angle doesn't change it'll still be 3x but right from going from sine to cos the 3 is going to pop out so it's going to be 3 sine 3x another thing is it's just not as simple as differentiation being the opposite of itself it's not so when you differentiate cos you actually get a negative sign you can also differentiate tan let's say we've got the tan of 4x when you differentiate it you get something called sec squared the angle is not going to change but the coefficient on the x is going to pop out so we're going to multiply by that if there's already a coefficient on this then that will be multiplied as well so let's say we've got 2 tan or 4x then when the 4 comes out we'll multiply the 2 to give us 8. sec squared is something you'll go through when you get onto the more advanced trigonometry by the way another thing you might see logarithms so if our function is ln x then when you differentiate that you actually get 1 over x you can also have all the numbers involved in this so let's say a function of x is 2 ln x cubed if you've got a power on the x that's going to come out as well so we differentiate it you would get 6 over x 2 times 3 equals the 6 and it was still going to be over x if you have an exponential so let's say we've got 4 to the power of x when you differentiate that you get somewhat of not answer so before the power of x stays which to be multiplied by ln 4. the last special case we'll look at is there's a number called e and e is an irrational number like pi i like pi screen a bit ease 2 in a bit but the decimal places don't repeat there's no pattern but they do go on forever now if you raise it to the power of x and you differentiate it something really interesting happens differentiating e doesn't change it it stays the same so effectively it kind of represents its own gradient now you can have all the numbers involving it so for example we could have 2 e to the power of 4x and when you integrate that what's going to happen is the coefficient on the x just like the fair trigonometry that's going to come out 4 times 2 is 8 but then the e part doesn't change it's still going to be e to the power of 4x so these are some special cases you'll see for differentiation some you'll see more than others the trigonometry ones are going to be the most common ones that you'll see if we have a curve let's say it's y is equal to x squared and we have a little look at what that might look like so we have an x and y axis we have the curve you might want to find something called tangents and normals so what do you mean by tangents and normals so a tangent is a line that comes down and it touches the curve at exactly one point now a normal is a similar concept but a normal is at a right angle to the tangent it also crossed the curve at that one point it might cross it at another point further back as well but again the point of this is it's at a right angle to the tangent so this is the normal and you might be asked to find the equation of these lines you'll also likely be given the coordinate of the point where this happens so let's save this question this wasn't the coordinate five two so let's look i would find the equations of the tangent and the normal so first thing to think about is that we've got our curve y equals x squared and if we differentiate it we're going to get the gradient of the curve or something that represents a gradient of the curve so we're going to reduce the power by 1 x squared becomes x to the power 1 i'm going to multiply the coefficient by the power so if there's no coefficient it's 1 1x squared 1 times 2 is 2. so we've got something that represents the gradient now because we've got a coordinate we can find the gradient at the point 5 2 so we can substitute in the x coordinate 5 and that will give us the gradient at that point 2 times 5 is 10. so we know at that point in that point only there is a gradient there's a steepness of 10. now this tells us something about the tangent the tangent has the same gradient as the curve at the point where the tangent is so the tangent is going to be y equals mx plus c and we know what the gradient is y equals 10x plus c all we want now is a value of c and to do that again i'm going to substitute in the coordinates at the point the five and the two so y is going to be equal to 2 then we've got the 10 lots of x and x is 5. and we've got the plus c which is what we're looking for this means that 2 is equal to 50 plus c so we take 50 away from both sides that means that negative 4k is equal to c so we see the tangent is y equals 10x minus 48. we can use a similar method for the normal so again for the normal we know it's going to be y equals mx plus c it's a straight line and we know what the gradient is the gradient is y equals 1 over 10x or our negative 1 over 10x plus c you might be thinking well where has the 1 over 10x come from and it's just the same as parallel and perpendicular lines so the normal is perpendicular to the tangent it's at a right angle so we take the negative reciprocal of it we can follow the same process to get c so we're substituting 5 2 so y is 2 the gradient's negative 10. we multiply that by x which is 5 and then we have the plus c that means 2 is equal to negative a half plus c because 1 over 10 times 5 is 5 over 10 which counts down to a half if we add a half to both sides that means that two and a half is equal to c so now we can write down the equation of the normal y is equal to negative one tenth of x plus 2 and a half so the big thing to remember here is that the tangent has the same gradient as the curve the normal being at a right angle to it is a negative reciprocal make it negative if it's a whole number put it over one if it's a fractured then you put the fraction upside down for the reciprocal we can use differentiation tells things about the turning point of a curve so let's say we've got the curve y equals x cubed minus two x squared plus four now if we draw a quick sketch of what this might look like you know a cubic is going to look something like this but we don't know exactly where this is going to fall on the x and the y axis and so this is what we're going to try and find out now before we even start there is one thing that we're going to know and that is with a plus 4 at the end it means that the y-intercept where the curve goes through the y-axis is going to be at 4. so we can find that straight away now if we differentiate twice so d-y by dx we'll get some information and we differentiate another time d2y by dx squared we get another set of information so let's do it so three times one is three which is a power by one two times two is four reduce the power by one and any whole numbers are gonna go away differentiating again three times two is six which used to power by one and it's just the next term we're just going to keep the coefficient so what information can we get from these well the first thing is we're probably going to want to know were the turning points of the curve up now to find those these are when the gradient is equal to zero and so what we can do is we say well if the gradient is three x squared minus four x we can make it equal to zero and then we can factorize it to find the those points now you might be able to factorize it by inspection or completing the square but you always be able to do it by using the quadratic formula i'm not going to go through that here because we've revised that already but if you do the quantity for this you're going to get that x is equal to zero and x is equal to one and a third so they are going to be our turning points so we know this will be one and a third now the other one's surprising it's a zero so really we should have had that a little bit further over we're not really sketch this right you can see it's not at x equals zero at all should be a little bit further over and what that means is that the y-intercept and one of the turning points are actually the same coordinate at 0 4. if you've got the y coordinate of the one and a third then what you can do is you can substitute it back into the y equals and that would give you the y coordinate as well if you wanted it another thing you might want is something that we call the point of inflection the point of inflection is in between the kind of two peaks so it's a bit where we start being the first curve and we start being the second curve now for that that's where the second differentiation is equal to zero the rate of change is equal to zero because it's in the middle of going up and going down so we're going to say that six x minus four is equal to zero and we can solve the equation we can add 4 to both sides we can divide by 6 x is 4 6 which would simplify to two thirds so we know that the point of inflection the bit where one curve turns into the other curve is going to be at two thirds another thing you might want to know is we've got the kind of the top and bottom of the two curves zero and one and a third we only know which one's at the top and which one's at the bottom because i've actually sketched out the graph correctly but you might not actually know which way around this cubic is so how do we know that zero's at the top and one third is at the bottom so what we do for this is again we'll substitute into six x minus four we have six lots of zero minus four and we have six lots of one and a third minus four so substituting the x coordinates of the top and the bottom into the second differentiation this is the rate of change and so what we want to do is look at what the rate of change is at these points now at zero six times zero zero take away fourth negative four we've got a negative rate of change up there what it means is at zero the rate of change is negative four that means the curve is starting to come downwards after it so we call this a maximum with the next one we've got six times one and a third which gives us eight and then we take away four and we get four it's a positive four that means that at one and a third the rate of change is four that means that the curve is going to start going upwards so we call this the minimum point it's these numbers about what's happening next when you read the graph from left to right so if on the right hand side it's minus four you must be the top most part that's going down next it's kind of the opposite way or anything if the rate of change is going to be four next then you must be the very very bottom for it to be going upwards afterwards one last thing is you know zero the top of that curve isn't actually the top of the graph because we can see on the right hand side actually goes up the cubic it goes down as well on the other side so these aren't actually the maximum minimum points we call them the local maximum and the local minimum because of the highest lowest points around about where they are but later on there are going to be even higher points and let's get this labeled on the graph so one and a third was the minimum point and zero was at the maximum point another thing you can do and it's not part of differentiation but if you take the original x cubed minus two x squared plus four if you do that equal to zero and you use the quadratic formula or again any other method you need to factorize this is going to give you the x intercepts now for this one there's only one and so it's going to give you x is equal to and it's roughly it's about negative 1.1 to one decimal place but often with cubics you're gonna get three of these and it's gonna cross all the x-axis up to three times so that's something else come up far as well so lots of different methods from different areas of maths but the key things are the differentiation will give you the turning points it'll tell you which one's local maximum and which one's a local minimum it'll also tell you the point of inflection which is the bit in between the two turning points differentiation usually works when you've got single terms are those terms will be separated by a plus or a minus if you've got single terms or multiple terms then you can differentiate now sometimes you might have in effect terms that are separated by other things so let's say they're being multiplied together now for that you're going to use the product rule so let's have a look an example first so you might have something like y equals and then you might have 2 x squared plus x multiplied by x cubed plus three x squared so you can see here you've kind of got two terms but to be multiplied together and so this is where you use product rule so product rule is if you're differentiating u multiplied by v then your answer is going to be equal to v d u by dx plus u dv by dx now i like to write that in a simpler way so i'll show that my way of writing it so we're saying if we want to differentiate u multiplied by v then what we're going to end up with is v multiplied by the differential of u plus u multiplied by the differential of v so what i want to do is we want to find these different bits when i find what u is we want to find what v is so u is going to be our first bracket 2x squared plus x and v is going to be our second bracket x cubed plus 3x squared and we can see with a little dash all that means is you differentiate it so u dash two times two is four reduce the power by one and the x is going to differentiate to one same thing with v three times was no number so the sum of three reduce the power by one three times two is six and reduce the power by one and so we've got our letters and we've got our differentiated version of the letters so now we just substitute so we want v times u dash so v is x cubed plus three x squared and we're gonna multiply that by u dash which is four x plus one we're gonna add to that u times v dash u is 2x squared plus x and v dash is three x squared plus six x now we just need to expand the brackets so we've got quadratics here in where the set up so x cubed times four x is going to be four x the power of four x cubed times one is going to be x cubed three x squared times four x is gonna be 12 x cubed and 3x squared times 1 will be 3 x squared moving on to the second bracket 2x squared times 3x squared 6x the power of 4. 2x squared times 6x is going to be 12x cubed x times three x squared is three x cubed and x times six x is six x squared so now we need to write down our final answer let's look at the power force first we've got four and six that makes ten ten x to the power of four then let's look at cubes so we've got one another twelve so that's thirteen another twelve so that's going to be twenty-five and another three that's going to be 28. so 28 x cubed then we'll look at squares we've got 3x squared and 6x squared that's 9x squared and so that is going to be our final answer 10x to the power of 4 plus 28 x cubed plus 9 x squared now this is probably one of the most complicated ones you can get with just kind of x's and powers usually it's going to be some plastic cancels out more when you do your differentiation you can also get this with things like trigonometry in and when you do things like that just have a look back at the special cases for differentiating trigonometry but it's going to work the same way the quotient rule is funny when i differentiate but the numbers are being added or subtracted like normal differentiation they're not being multiplied like product rule they're being divided so you might have something like y equals 2 x cubed plus 3 x squared all divided by eight x plus five now to use quotient rule what you need to do is you wanna differentiate you've got u divided by v or something b divided by something else and so the answer is going to be v multiplied by d u by dx take away u multiplied by dv by dx and it's all going to be divided by v squared now again a little bit of complicated notation that's why i like to write it out and a little bit a little bit more simple so let's look at my way so i would say if you've got u over v and you want to differentiate them then you're gonna do v multiplied by the differential of u take away u multiply the differential of v all divided by v squared so first let's figure out what u and v are so u is going to be the numerator 2 x cubed plus 3 x squared and v is the denominator 8x plus 5. so u dash we are going to differentiate the numerator 2 times 3 is 6 which is the power by 1. 3 times 2 is 6 due to power by 1 same thing for v 8 x becomes 8 and the 5 is going to go away and now we can substitute so we want to do v multiplied by u dash so v is going to be eight x plus five and u dash is six x squared plus six x so that's what these in brackets have been multiplied together we take away from that u which is two x cubed plus three x squared multiplied by v dash which is a and it's all gonna be divided by v squared which is 8x plus 5 multiplied by itself it's squared so now we need to expand all of the brackets so 8x times 6x squared it's going to be 48 x cubed eight x times six x is going to be rtax squared five times six x squared is going to be 30x squared and five times six x will be 30x we're taking away from that 2x cubed times 8 will be 16x cubed and 3x squared times 8 is gonna be 24 x squared remember we're taking away the second bracket so these are going to end up being negative that's all going to be divided by we've got the quadratic to expand 8x times 8x is 64 x squared 5 times 8x is going to give us 40x and 5 times 8x is another 40x and 5 times 5 is 25. so i'm going to collect like terms and try and simplify this but the quotient rule this pack is done now it's just kind of tidying it up let's look at the x cubes first we've got 48 x cubed and then we've got a negative 16 x cubed so 48 take away 16 is going to give us 32. then we look at the x squared so we've got 48 we're going to add on 30. i'm going to take away 24. that's going to give us 54 x squared then we've got the x term there's only one 30x then the denominator we've got 64 x squared 40 plus 4 equals 80 x and the 25. now at this point there's things you can do so what you might want to try is you could try and factorize this and you've got the same factor of the numerator and the denominator then you can divide by and make it simpler you might just be able to look at the whole thing and go over every term for an x here i can divide 3 by x or every single coefficient is even so we can divide 3 by 2. so there's things you can do at this point on the words but then that's outside the quotient rule it's just a simplifying algebra so i'll leave it here the chain rule is when we differentiate and different bits of your expression and it's separated by plus or minus with for normal differentiation they're not separated by multiplications for product rule or divisions for quotient rule the chain rule was going to be some kind of other function so it could be a power or it could be a bit trigonometry or something so for example you might have y is equal to and you could have something like 4 x cubed plus let's say 3x squared and this itself is going to be raised to a power so this could be to the power of let's say five so to use a chain rule what you say is d y by d x is equal to d y by d u multiplied by d u by d x you might have to know what's this u bit about well it refers to some substitution now it might be easier to write this in a different way so if you've got a function of x and x is another function of x then we can change that to b the differentiated version of that function that is a function of g of x multiplied by the differentiated version of g of x and again that still looks a little bit complicated so let's have a look at it in practice and see if it makes more sense so our function of x is what we are actually doing to this equation and it's being raised the power of five so it's gonna be something to the power of five now rather than saying oh it's something to the power of five we're gonna actually use a letter for this and we're gonna use u that's where the u came from before and we put that into the x so our function four x cubed plus three x squared it's being raised to the power of five now what's inside the brackets what's been you know what the function has been applied to is our g of x and that's 4x cubed plus 3x squared so we've separated what's inside the brackets and what's in outside the brackets we've done this with a power in this situation you can also have this with trigonometry so when you kind of differentiate the angle in trigonometry then the f something it might be sine of cos rather than a power five with the x bit being the bit inside the trigonometry brackets so now we need to differentiate so differentiating f of u so we know we're gonna reduce the power by one so five it's gonna become four and we multiply by the old power so we get five lots of u to the power of four then we differentiate g of x so four times three is twelve reduce the power by one three times two is six reduce the power by one i've got the differentiated version of g so now we just need to substitute so what they say is that it's the differentiated version of f which was five of our function the power of four we're going to put g of x back into that which was four x cubed plus three x squared then we're going to multiply that by the differentiated version of g which was 12x squared plus 6x multiple terms first let's put in brackets so for our final answer we're trying this up we've got the whole number 5 at the front we've got one set of this bracket and we've got four sets of this bracket so we've applied chain rule here we've still got quite a complicated expression at the end there might be more steps here you might want to take more things out as factors we might want to expand the brackets it all depends on what question you've been asked but this is the chain rule part of the question integration is the reverse of differentiation so let's say we have y equals 2x squared plus three x plus four if we were to differentiate this to get the gradient then we'd multiply the coefficient by the power two times two is four and reduce the power by one anything in x we're just to have the coefficient left over and any numbers are going to go away so we get 4x plus 3 is the gradient of 2x squared plus 3x plus 4. so we have differentiated so how do we integrate how can we go in the opposite direction so we think about differentiation there's a couple of features first thing is the power goes down by one and then you multiply by the old power so integration is going to be the opposite you increase the power by 1 and you divide by the new power and we can write a general version of this so if we integrate this is the integral symbol if we integrate let's say a x to the power of b then we put dx at the end as well which is part of the notation for this then the answer you increase the power by one so b becomes b plus one and then you divide by the new power set a is going to be divided by b plus one so yeah differentiation take one away from the power integration add one to the power differentiation multiply by the old power integration divide by the new power so let's give it a go let's do this let's see what this does so if we're going to integrate 4x plus 3 then what we do is we're going to increase the power by 1 so the x becomes x squared and then we need to divide by the new power so 4 is divided by the 2 to get us 2x squared then for the 3 we increase the power if there's no x there then it's x 0 increase it by 1 we get x to the power of 1. and then we divide by the new power while the new powers are one three divided by one is three and that's it and now if you compare this to differentiation we're actually missing the plus four so when we work backwards it isn't actually a way to get that plus four back so what we do is we write plus c plus a constant there might have been a number but we can't actually work backwards to find what it is so there is a flaw with integration now one way to make up for this is something called a definite integral what we just did there is a indefinite integral so if we're going to integrate o 4x plus 3 what you might be given are numbers on the integral and let's have a look what this means so we're going to do the integration which we already know is 2x squared plus 3x plus so what this means is you're actually substituting those numbers so the first term goes to 10 is the 3. that means we have two lots of three squared plus three lots of three plus c then we take away substituting two lots of one plus three lots of one plus c and then we can work this out so three squared is nine times two is eighteen three times three is nine a plus eight then we take away the second bracket one squared is one times two is two we're taking the second bracket away so we take that away three times one is three but again we're taking away and then we've got the plus c and again because we're taking away the second bracket we have to take away the plus c and if you do that we'll notice is we've got a positive plus c and a negative plus c and they're gonna cancel out so you can actually solve a definite integral even if you don't know what c is so we can add these up 18 plus nine take away two take away three and all together we get 22 so we get a full answer without needing to know every single bit of the thing that we integrated now you might be wondering what does this represent so let's try it on a graph and see what it actually have x and y coordinates we are going to have the curve which is our 2x squared plus 3x plus 4. so what does the definite integral mean well we saw two things three and one so what this means is you have one on the x-axis and you're gonna have three on the x-axis so this is what we're looking in between and what happens is when you integrate what you actually do is you're finding the area underneath the curve between the curve and the x-axis so everything i'm highlighting in blue this is our answer that's what we're looking for and that is what we discovered was 22 so if we label it this has an area of 22. this then takes another function of integration as well as being the opposite differentiation and so you can turn a gradient back into the line like we did with our example you can also use it to find the area between the curve and the x-axis if you integrate something that's y equals now there are special cases for integration and it's based the same with differentiation but the other way around let's start off with trigonometry so if we have f dash x which is cos of x then when we integrate it to get f of x when you integrate cos you get sine works the same way that it does with differentiation now you can also have numbers on the x let's have a look at so let's say we have cos of 2x then when you integrate it you're going to get start to x again the angle is not going to change but with differentiation you'll be multiplying by the two well integration's the opposite you need to be dividing by the two so you end up with a half of sine 2x if you do that you can integrate sine so let's look at sine of 3x so when you do this you remember differentiation when you differentiated cos you got a negative sign so this is the opposite of that so if you integrate sine you're going to get a negative cos again the angle is not going to change we could divide by the number it's a 3 so we are dividing by three then we do the same thing with tan now with tan because the opposite will start off with sec squared and let's say we've already got a four on the outside so when you integrate it when you integrate sec squared you get tan the angle doesn't change but you divide by the angle and we've already got a 4 on the outside so 4 divided by 4 is 1. so we just get tan of 4x we also have the reverse of differentiating logs so f of x is one over x then when you integrate it you're gonna get ln x if f dash x was something like five over x then f of x is going to be ln x to the power of five we also have e if f dash x is equal to e to the power of x then when you integrate it you can ease that special number where when you differentiate data it stays the same and it's the same for integration however if you have special numbers involved in it they can get involved so for example e to the 3x when you integrate it you're actually going to get a third of e to the 3x so the e pair hasn't changed the expert hasn't changed but you bring out the three if it's differentiating you multiply by it it's integration you divide by it another useful thing to show is you can have things like you can have 2 over the square root of x now there are any special rules for this it's just a matter of rewriting this in a more reasonable form so we can say well this equals 2 over x to the power of half square root is the same as a half x per half and then we can bring that up to the top until it's 2 to the power of x the negative half so any powers on the denominator we brought up to the numerator now we've got it written like this now we can integrate it safely we just use our normal rules we increase the power by one so x to the negative half becomes x to the positive half we've added one and then we divide by our new power so it's gonna be two divided by a half required as a fraction for example you write it as a decimal two divided by half how many halves are in two there are four halves and two so you can get some tricky things going on in that situation another way to divide by a half is to multiply by its reciprocal flip the fraction upside down you get 2 over 1 which is 2 so 2 times 2 gives you the 4 as well so you're trying to integrate you've got some really free notation you've got fractions you've got roots try and change them into a single line like i've done here and then it should be able to integrate integration by parts is when the different terms that you're integrating and separate by plus or minus maybe they're being multiplied together so for example let's say we've got y is equal to i would have something like x minus 3 and we could have x squared plus 5x so how would we integrate this so using integration by parts a simple way to write it your integral if you've got u multiplied by p then that's going to equal u times v take away the integral of v times u so it looks pretty complicated let's try and identify what u and v are so u is gonna be one of our brackets let's just say it's the first bracket x minus three so we can differentiate it to get u dash because we need that's part of our formula which would be one v v dash that's how it is in the formula is going to be our second bracket that's x squared plus five x so if we integrate it to get v we're going to increase the powers by one and divide by the new power so we should get something like this so now we can substitute so for our answer we're gonna have u times v that's gonna be x minus multiplied by a third of x cubed plus 5 halves of x squared take away from it the integral of v times u dash so v is a third x cubed plus five over two x squared and then u dash is once we multiply it by one it's not going to change and so now we need to integrate that for our final answer so the first pair isn't gonna change here but this second part is so we need to increase the powers by one again so x to the power of three becomes x to the power of four we need to divide by four we're already dividing by three divided by three and four times together we get twelve then we're taking away that so put take away sign in we ordered the second term x squared is going to become x cubed 3 divided by the 3 we're all dividing by 2 so 2 times 3 is 6 and we're taking that away and so now we've got our full expression for integration by paths now there's things we can do here so you can expand the brackets you could try and factorize things there's other options from this point onwards but what we've got here is the kind of essence i integrate by parts and yes there might be then some manipulated algebra afterwards last thing to remember that when you do this we're going to get the plus c because we're integrating so we've got to include the plus c we also have integration by a substitution this is when normal integration you terms separate by plus or minus if there's multiplication involved perhaps brackets next to each other integration by parts if you have powers it's going to be integration by substitution so you have something like y equals then we could have three x plus two cubed and we want to integrate this so here's how we do it so if the integral if we have a function of g x multiplied by g dash x then we can say that that's equal to the integral of the function of u so the f part i'm going to call it f of u because i'm going to take out what's inside the brackets as a substitution so that's going to be a bracket and it's cubed then we've got our g of x that's what's inside the bracket it's three x plus two and this is also the same as u now we need the form of g dash x so if we differentiate it we're going to get a 3. now what this formula means is you can only use a shortcut and change that kind of complicated bit on the left to the simpler bit f of u on the right you can only do that if you've got g x in here and we don't we don't have a 3 on the outside of this bracket so we need to put it in so if we go back to our original statement we want to integrate 3x plus 2 cubed and we can only use integration by substitution if we are multiplying this by g dash x which is three so we're going to put it in so get our integral we've got three x plus two cubed so really you wanna multiply it by that three now are we allowed to do that well there's a way we can do it because obviously if you multiply by three we've got something three times bigger and it's going to mess up the answer if we also put a divide by three in here it's now going to work because we've got to cancel each other out and now we can use the integral of f of u so that's what we're going to do next so the integral of the function of u we've actually done the function of u so we can substitute that in and it was u cubed so if we differentiate u cubed we need to increase the power so the 3 becomes a 4 and then we're going to divide by the new power and of course we get plus c because we integrated then we can substitute u back in u was equal to g of x which was over three x plus two we also need to remember that we had the third part of this we're gonna put third in that means we've got a third multiplied by a quarter if you multiply those together you actually get 1 12. so our final answer would be a 12 of 3x plus 2 to the power of 4 plus c and again at this point there might be brackets to expand there might be factors to take out there might be more steps but we've done the integration using substitution integration is to find the area under a curve it's particularly between a curve or a line and the x-axis now look at the diagram here it's not between a curve or a line and the x-axis is in between a curve and a line so how can we integrate this now it's just as simple as combining the two lines together because both lines are equal to y so x squared plus 12x plus 32 is equal to y and x plus 8 is equal to y so they're both equal to y then they're both equal to each other so x squared plus 12x plus 32 is equal to x plus 8. now what we could do now is combine this on to one side so we can take x away from both sides 12 to the equals 11x and we can take 8 away from both sides 32 take away 8 is 24. and now they're combined together now we can integrate that so we can integrate x squared plus 11x plus 24. and if we integrate this combined line and it's going to give us the area in between the two lines so to integrate it we are going to increase the power so x squared becomes x cubed and divide by the new power x becomes x squared divided by the new power and 24 there's no x so we're going to put an x in so now we've got something that represents the area normally to do now for this expression is to use a definite integral and it's going to be in between the x coordinates were given because we look at these negative three you look at the graph is the end of this curve and negative a is the star of this curve or at least there's the shaded area so we do the definite integral so we're going to substitute the largest number first negative three that's going to give us a third of negative three cubed plus eleven halves of negative three squared plus 24 lots of negative three now i could write the plus c down but the next step is going to take away the substitution for a smaller number which is negative eight and so we're taking away we get a negative c out of this so the c's are gonna cancel out so we can we can safely ignore them so we're gonna have a third of negative 8 cubed plus 11 halves of negative 8 squared plus 24 lots of negative 8. and if we work all this out then we're going to get an answer for the area so negative three cubed is negative 27 and then divide it by three because what a third of it that's going to give us negative nine negative three squared is nine what eleven halves of its times by eleven divided by two that's going to give us 49 and a half they want 24 times negative 3 that's going to give us negative 72. moving on to the second bracket we're going to take this bracket away negative 8 cubed is going to give us negative 512. i want a third of that so we divide by three that's going to give us negative 170 and two-thirds say 0.6 recurring now we need to take this away it's already negative so that'll give us a double negative it'll be positive then negative 8 squared is 64. what 11 halves of it times by 11 divided by 2 that's going to give us a positive 352. we're taking away this so we're going to take it away then we've got 24 times negative 8 it's going to give us negative 192 and again taking this bracket away will be double negative so we're adding it so we can add all this together for our final answer negative nine plus 49.5 minus 72 plus 170.6 recurring take away 352 and add 192. all together that's going to give us a negative 20.8 to one decimal place now how can we have a negative area well we can so when you get negative area you can just assume it's positive so you can ignore the negative sign so going back to the graph we're saying the green highlighted area we've got 20.8 squares shaded in to one decimal place you can also use this method find the area under a curve for when you have a quadratic and a cubic and the intersection creates two separate areas which i'll label got area number one and area number two there's a similar method but there's one important difference part way through now i need a lot of space for this method so i'm gonna try and compress combining the two things together so i'm going to use the cubic as a base we've got x cubed now with the squared bit negative 10x squared we've also got a positive x squared in the other one so we need to take away x squared to combine them so it'll be negative 11x squared then we've got the 29x and we're taking away 5x on the other side we have to add 5x to both sides to get rid of that so that would be 34x then with the numbers we've got negative 20 plus 4 on either side take 4 away from both sides to give us negative 24. so this is what we're going to be integrating so if we integrate it increase all the powers by 1 x to the power of 3 becomes x to the power 4 and divide by the new power x to the power 2 because x to the power of 3 divided by the new power x becomes x squared now dividing 34 by 2 that should give us 17 and the 24 to become 24 x and as usual gravity plus c now what's different here is what the definite integral is now look at the x coordinates got six four and one what we're going to do is we definitely integral for each cherry one at a time so for the first area we can see it's in between the one and the four so the definite integral for this is going to be we're substituting the largest one which is the four so we're gonna have a quarter of four to the power four take away eleven thirds of four cubed plus 17 lots of four squared minus 24 lots of four they take away from that second number which was a one so gonna have a quarter of one to the power of four take away eleven thirds of one to the power three plus 17 lots of one squared minus 24 lots of one so there's a lot to do there but we need to get a number for the first area so we have four to the power of four and then a quarter of it divided by four will give us 64. four to the power of three eleven thirds times by eleven divided by three is going to give us 234.6 recurring four squared is sixty times it by 17 is going to give us 272. then we're going to take away 24 times 4 which is 96. moving off the second bracket and remember that we take this one away so we're substituting one and do a power to one we're still gonna have one so a quarter of one that's when he was negative a quarter 11 thirds it's already negative we're taking it away it's gonna be double negative then we've got the 17 positive we need to take it away it'll be negative and then at the end we'll go to 24 so it's already negative we're taking away that's going to become a positive so we can add all this together to get the first area so 64. take away 234.6 recurring plus 272. take away 96 take away a quarter 1.25 then we're going to add 11 thirds so add 11 over 3 take away 17 and add 24 all together for this and get in and it's 15 15.75 or 15 and three quarters now what we need to do now is what we found though that's only the first area we need to find the second area and so we're going to do another substitution so the second area is in between four and six you look up to the top that's the end point so the largest one is six little shooted quarter is six to the power of four take weight eleven thirds of six cubes plus seventeen times six squared take away 24 times 6. then we're gonna take away the smallest one which is the four now we've already substituted in the four so we can already get a value for this and it was the first four terms so it was the 64. take away 234.6 recurring plus 272 take away 96. so that actually was five and a quarter so we've already got the value for that so all we need to do now is our third substitution for each six so we have a six to the power of four what a quarter of it divide by four that's 324. six to the power of three what eleven thirds of it times by eleven divided by three that's going to give us seven hundred and ninety two six squares thirty-six times it by seventeen that's giving us six hundred and twelve and then twenty-four times six is going to give us 144. we're going to take away from that the five and a third and all together that should give us the second area so 324 take away 792 plus 612 take away 104 by four take away five and a third that's giving us negative five and a third isn't it odd we get that same number twice basically everything when we're substituting the six seems to have cancelled itself out so for areas we only take the positive so it's five and a third that's the second area so now if you want the total area we just need to add those together so 15 and three quarters plus five and a third that's going to give us so you can write it as a decimal it'd be 21.1 to one decimal place over an accurate answer you could write it as a fraction it's 21 and 1 12 for an accurate unrounded answer to the curve here the biggest source of error is that that second area is listed as negative but you treat it as a positive area areas are always positive so make sure whatever numbers you've got you make a positive and you add them together at the end vectors give a direction and they give a length so on the first diagram we can go from a to b and we can see it's labeled vector a so we have two a's there's the capital letters at the point so you can see the coordinates the positions of vertices whatever you want to call them and a with a lowercase a is the vector so we can say the vector from a to b is equal to a now the thing with vectors is they're not limited to particular positions so for example from c to d you'll notice that's the same length and it's the same direction as a to b so what we can do is we can say well c to d is also equal to a you'll also see vector b so from a to e we can call that vector b and what we can do as well is we can go backwards so we can go let's say from m to i in the opposite direction to b but you'll see other than it being going in the opposite direction it's otherwise the same direction and the same length and so what you can do is we can say that the vector m to i so the direction is really important we're not saying i to m let's play m2i we can say this is negative b it's going in the opposite direction another thing we can do is use multiples so for example we could go from f to h and we'll see from f to h we've gone the same direction as a and we've not gone the same length we've gone twice as much the length is twice the length of a so we can say from f to g is equal to two a so we can do maths with the lengths we do the same thing with b so let's say we go from g to o then vector g o would be equal you see again same direction as b but it's twice as long so we call it t b if you want to go the opposite direction and go from o to g then we call it negative 2b another thing we can do is we can go halfway so we could go from let's say n halfway over to o so we can see it's the same direction as a but it's half the size so the vector from m to o is equal to half of a you can also do combinations of vectors let's say we wanted to go from let's say from m to j now there's no vector labeled for that direction what we could do is go from m to i which is already labeled and then go from i to j and we can go the long way around and this is the loading vectors then we're saying the vector from m to j is equal to so it was a negative b from m to i and a positive a from i to j so we can combine vectors together as well our second diagram has got different vectors it's got a different layout so it's still from a to b it's still going to be vector a so that's the same for this diagram just by coincidence but if we have a look at vector b you'll see that this one's different it's a slightly different direction as well we go from a to o this time that's vector b a to o in the previous diagram would be you know something completely different so just like algebra the value of the letters can change depending on the question you've got they're not always constant now you'll notice we have another direction we can go in and go from b to o and from b to o there's no labeled vector so we can't immediately label anything we can't just say oh we'll call it vector c you're not allowed to do that you can only use what you've been given so what can we say about this well we should go the long way around we could go from b to a which is the wrong way down vector a to negative a and then we can go from a to o which is the right way down vector b so we're adding b so again with vectors you can go the long way around if you need to you'll also notice if we for example went from b to c that's plus b and c to o is negative a so even if we went a different direction we would still get exactly the same answer so now we know what beta always which is negative a plus b we can stare to use it to find other vectors so let's say you wanted to go from c to e the direct route doesn't have any lines labeled so we can't take a direct route but what we could do if we go from c to d and we can go from d to e and that'll get us to the same endpoint now c to d is the same direction and the same length as b to o we could even write this with a little equation so we could say that c to e is equal to c to d plus d to e so we know that c is d being same as b to o will be negative a plus b and we know that d to e it's the opposite direction of a so then all together that's going to give us negative 2 a plus b so you can start to do maths with vectors as well you'll also commonly see positive vectors put first so minus 2a plus b you might see b minus 2a it represents exactly the same thing but putting a positive number at the front just means that you don't need to write the positive sign down because if there's no symbol at the front we assume it's positive so we're just saving one space by writing it that way around you don't get extra marks for it but you might see it written that way in math schemes now there's multiple ways of writing vectors so for example here we've got vector p and we've got vector q and so just using the single letters is just like what we looked at on the last revision slide we can split them apart into i and j vectors and these give us a bit more of an idea about what the direction of the vector is like so let's sketch out what these things mean so i correspond to the x coordinate and y corresponds to j so when i say 10 i what you mean is that we are going across by five when we say for j we mean that we're going up by four now it doesn't mean go across and then go up what it means is that you're gonna go directly in between the two points but rather describing it as a diagonal line we describe it as a horizontal and a vertical component so we've got vector p and we've got the measurements of vector p which is what 10 i and four j basically give us now what about vector q so vector q we can see it's four i so it's going to go another 4 across and it's negative 9j so it's going to go 9 down so the plus and minus just tell you it's going left or right or up or down so now we can label vector q so it's going to go directly it's not going to go left then down so it's going to look like this so we've got vector p and at the end of vector p we've got vector q now we can do more math with this so for example we could have vector p plus vector q and so to do that we take vector p 10i plus 4j and we take vector q 4i minus 9j and i'm going to add those vectors together so 10i plus 4i is 14i and 4j take away 9j is negative 5 j now we can also show this vector on the diagram so p plus q we're going to start at the start of p i'm going to end at the end of q so again it's not like a complicated wave of instructions we do 10 different instructions you go right and then up and then right and then down what you're actually doing is you're going directly from the start of p to the end of q and so p plus q is what i'm labeling in pink so again there's a distinction between what the vector actually does which is a direct root and the way it's written which might be giving you vertical and horizontal measurements to give you an idea of where the vector actually points now you can also do multiples of these let me do the previous vectors so if p plus q is equal to 14i plus five j then we can say that two lots of p plus two lots of q you know that's going to be something like 2 lots of 4i minus 5j in fact you might actually write the original expression using brackets as well so you can write them out in ways like this you can expand the brackets as well so the bracket is a good way to show what you're going to do so 2 times 14 is 28 2 times 5 is 10 and so 2 lots of people's q would be 28 i minus 10 j if we were to draw that out then what would happen is we would get something like this and so it would extend out further on it would be twice as long now i and j vectors are called position vectors and we can use them to draw shapes and also prove things about shapes so we're looking at a parallelogram let's try and draw the parallelogram first so these position vectors they all come from the origin they come from you know the point we're measuring everything from so let's call the origin o and let's try and draw all the different vectors so that to a it's going to go minus 2i so 2 to the left and plus 4j so 4 up so here is where point a is going to be so the position vectors have just given us an idea of the direction that vector a goes next we'll look at vector b so it's 8 i so 8 to the right and 3 j so 3 up again the vector just gives us the direct root so two components just give us the length of it in both directions then let's look at c so c is negative i and then it's plus 10 j so it's going to be something like here and again it's giving us the direct root the last one for d so it's minus 11 i and plus 11 j so it's going to be somewhere around about here and so that's going all the way over here so we can see it looks like a bit of mess at the moment so we've gone from the origin in four different directions and we're saying this is a parallelogram so let's see if it does make something that roughly looks like a parallelogram so we're going to go from a to b from b to c c to d and a to d and we can see you know it could be possible that this is going to be a parallelogram now i've sketched this out i have i don't have a grid here that you can't see so it is kind of vaguely accurate as well but you can get the idea of where all the different vectors are going to see and how it's describing where the shape comes from the next step is how do we prove that it's a parallelogram so for the proof what we need to do is we need to show that the opposite sides are equal so a to b should be the same vector as c to d or it's going to be in the same order so from d to c so i'm just going from left to right here so be careful which way around the letters are so if the same length and the same direction then they must be parallel the same thing to the two sides so we need to have a today would have to be the same as b to c so we need to look at what these vectors are so let's look at a to b first so to go from a to b we can't go the direct route there's no vector label there but there is a vector from a to o that was negative a and there is a vector from o to b that was positive b so if we had the labels now we're going negative donate and then we're going positive at b we also had vector c and vector d as well which we'll need later so what's negative a going to be well we've already got a is negative 2i plus 4j so the negative version of that we're gonna flip the signs so it's gonna be t i minus four j then when i add b on to that which is a i plus three j so if we add all this up plus like terms two i plus eight i is 10i negative 4j plus 3j is negative 1j so that's what a b is so now let's have a look to see if it's the same direction and the same length as d to c so to start at d and end at c we can't take the direct route but we can go via the origin because that's what the position vectors are giving us so we can go backwards down d to get to the origin and then go forwards up c to get to c negative d flip the signs on d it's going to be 11i minus 11j and then c what a positive c so it's the same one it's written down negative i plus 10 j so collect like terms again 11i minus i is 10 i and negative 11 j plus 10 j is going to be negative 1 j and what you can see here is that the vectors are the same they're both 10 i minus j so if a b and d c are equal to the same thing that means therefore they are parallel so we've proven one set of parallel sides and we can use a parallel notation on this as well now these are definitely parallel and they're also the same length so what we need to do now is follow the same process to see if a to d and b to c are also the same length now i'm not going to do that because i think we revise this enough now so you should know what to do next so you can either call that yourself and you should be able to get the same answer for a d and b c but it is actually a parallelogram saying prove it's a parallelogram that it is going to be a parallelogram if it says something like is it a parallelogram then it not being one might be an option and then you'd prove it by showing that these answers are different just be careful with the wording of the question because you made a mistake here and they didn't appear to be parallel and it's saying prove it's parallel you should know that you've made a mistake you need to go back over you're working out another thing to do with vectors is co-linear lines all this means is that it's a straight line so again we can try and sketch this out if you've got a grid you can do this accurately so starting from the origin a is one to the right and two up b is seven to the right and it's two down and c is ten to the right and it's four down so we have the vectors for these o to a remits a direct route o to b and o to c now looking at a b and c themselves if they're called linear then they're going to be a straight line so let's see if they actually are a straight line so joining up a all the way over to c we should have b stop off in the middle and we can see since i've drawn this accurately you can see they are actually all on a straight line however you're not always going to have some graph papers do this accurately on so we need another way to prove that they're all on a straight line and then it's going to be by looking at the values of the vectors so what we're going to do is we want to look at vector a b and we want to look at vector bc and there should be something about the two different sections of the line that tells you that they're straight so let's look at a b first so to start at a and end at b and there's no vector in between a and b but we could go from a to o which would be minus a and we go from o to b which is plus b the negative vector of a is negative i minus two j and the positive vector of b is seven i minus two j add up the i's you get six i and add up the j's you get negative four j now let's look at b c and c is there anything we can convert so b to c we can't go direct but we could go backwards down b to the origin and then forwards up to c to get to c negative b is negative seven i plus 2j and positive c is 10i minus 4j so collecting like terms negative 7i plus 10i is going to be 3 i then looking at the js 2j minus 4j is going to be negative 2 j so look at the two answers they do appear to be different so how can we say they're on a straight line when the vectors are different well what we can do is you can try factorizing them now if we factorize a b we can take two out of both and that will give us three i minus two j and that's actually the same answer that we had for bc so we can see the actual inj bit is the same length and the same direction it's just that a b is multiplied by two as well which means that a b is twice as long as bc so we're proving a straight line they don't need to be the same direction and the same length a straight line they're only going to be in the same direction and so you all need to look at the direction bit the directions don't match up have a look see if you can factorize to make the directions match up the only real way that is you can't have any bits of ion j outside of the bracket the bracket bit has got to be exactly the same at the end we can just write a b and b c have three i minus two j as a factor therefore they're a straight line they're called linear position vectors give a vector in terms of a kind of x and y length you can always have vectors written like this diagram where we have the length of the vector and we have an angle of the vector now one thing you might want to do is you might want to change this into a position vector so it might be we want to know how far across is which is going to be our i vector i want to know how far up it goes which is going to be our j vector so how do we do it well you might notice you might be able to draw this as a triangle so we have a hypotenuse 11.7 so that's the total length of the vector we want to get j and i so we can write i here and we could write j on the other side of the triangle it's going to be the same length we also know the angle this is at it's 42 degrees now given this is going to be a right angle because on a coordinate grid you know i and j are at right angles to each other like a x and y axis are then we can use trigonometry for this we can label the sides again we've got the hypotenuse got the adjacent and we've got an opposite so we can use sokatoa we're going to want a formula triangle for this and we can't find them both at once we can only find one at a time so let's try and find the i first i realized label on both eye let's make sure the one in the wide rection is going to be a j so with i that means we've got the adjacent and we've got the hypotenuse as a clue so it's going to be a car question so write can form the triangle and we'll look in for what the adjacent is so if we don't have the adjacent we've only got the cosine and the hypotenuse so we need to multiply them together so it's going to be the cosine of the angle 42 degrees multiplied by the hypotenuse 11.7 and if we do that calculation we're going to get the kind of distance to the left that i goes so cos 42 multiplied by 11.7 gives us eight point six nine to two decimal places now let's have a look at j so again glue the same thing we need a formula triangle for j look at the j that's the opposite we've got the hypotenuse of the clue it's going to be a solar question we're looking for the opposite so it's going to be sine times the hypotenuse so sine 42 times 11.7 now if you just look at two calculation notes are quite similar so you can use your previous calculations on the calculator screen by pressing the back button and swapping cos the sign what you should get is 7.83 to two decimal places so now we can write this out as a position vector so i is 8.69 and j is 7.83 now the last thing to do here is either going to be positive or negative because it all depends on where the angle is measured from now if the angle starts measuring from the positive x side then it's going to give you its negative or positive already so you don't need to think about it if it's from the negative x-axis and it doesn't go anti-clockwise it's going to go maybe clockwise from here onto clockwise from here then we need to be looking ourselves if it's positive or negative so out of the four possible directions you can measure from from the x-axis if you're going from the positive side anti-clockwise it will give you a positive or negative if it's going in the other other directions then you need to be looking if it's positive or negative yourself if the angle is measured from the y-axis then what you need to do is take it away from 90 to kind of make it so you've got the the kind of equivalent angle from the x-axis these have to be measured from the x-axis now think about the quadrants of one of these x is positive on the right-hand side and it's negative on the left-hand side y is positive at the top and it's negative at the bottom so you just look at which the four quadrants are in and we can see the quadrant we're in is the top left which means that x is going to be negative so i will be negative and j is going to be positive so our final answer for the position vectors is negative 8.69 i plus 7.83 j what we have here is a column vector and you can rewrite it as a position vector it would be 8.23 i plus 11.33 j so with a column vector the top one is x the bottom y is y x is i and y is j now what if you wanted this to be a kind of total length and an angle like the last bit we looked at how do we know the angle of this direction how can we find it so again it's going to be about triangles so you can imagine we've got the i vector or the i pair of the vector 8.23 we've got the j part 11.33 and so the total length of the vector is going to be from the start of i to the end of j and we call this the magnitude which we label r we're also going to want the angle of this which we're going to call theta so how do we find these two things as really easy we've got two sides of triangle what the hypotenuse of a triangle it's pythagoras theorem so we're going to square the two sides we're going to add the answers together and then we're going to square root that answer to get the hypotenuse so you can type all of that into calculator on one line i'm going to do it in two actually so i'm going to write out the squared bits and add them and then i'm going to square root the answer you'll write it in one line you write square root first and then you put brackets everyday everything that's been square rooted when you do that you should get the answer which is 14. now it's 40.00 so to one or two decimal places it's going to be 14. so that's the magnitude of the vector it's the length of the vector so we've got an x length a y length and then the total combined lengths so next what about the angle how do we find the angle now again it's going to be using trigonometry so we've got the hypotenuses are the adjacent is 8.23 and the opposite is 11.33 so we're going to need a formula triangle we're going to use sockatoa now we have o and a as clues so it's going to be a torah question and we're looking for the angle so we can see we've got o over a that means that the term of the angle is equal to the opposite 11.33 over the adjacent eight point two three so solve the equation it needs to do ten to the minus one to both sides to get theta on its own it's gonna be ten to the minus one eleven point three three over eight point two three that can be types into calculator remember to open the brackets that's really important divide the numbers by each other close the bracket and you should get 54 degrees and that's actually 54 to 1 decimal place as well so now we know from the column vector 8.23 over 11.33 we know the total length of the vector the magnitude of the vector which is 14 and we know the direction it's going in is going at 54 degrees from the x axis ouch this is why in some videos i have unexplained scratches

Channel: Science and Maths by Primrose Kitten

Views: 909,671

Keywords: a-level, aqa, edexcel, ocr, cie, as, a2, exam, revision, a level maths, alevel maths, a level maths revision, Laws of indices, Surds, Expand brackets, Factorise quadratics, Simultaneous Equations, Factorise Cubics, Algebraic long division, Functions, Partial fractions, Plot linear, Solve inequalities graphically, Solve inequalities algebraically, Transform functions, Equation of a circle, Parametric equations, Binomial expansion, Arithmetic progressions, Geometric progressions, Radians

Id: sOE8Slo3Pqw

Length: 234min 2sec (14042 seconds)

Published: Sat Oct 03 2020