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    Hello, I'm currently trying to battle my way through the sketching graphs topic in my edexcel c1 textbook but i've tried doing this exercise for hours but i really can't do it, so I was hoping for a little assistance?



    Basically, if I know the method to the do first couple of questions then I think I could manage to do the rest but I just can't get the answers for what I'm looking for.

    The first question I know that the co ordinates for 2 out of 3 intersections as:

    (0,0) and (4,0), I tried using the method: x^2(x-4)=x(4-x) and then cancelling it down but I just got a really silly answer. I looked at the answer booklet and the remaning co ordinate for intersection is (-1,-5) but i'm totally stumped on how to get to this
    Basically I did:

    x^3-4x^2=4x-x^2
    then i cancelled it down to..
    x^5-4x^2-4x which simplified durther gives x^5-8x^3...

    Same with the second one(3b), the equation I got was 5x-x^4 and I have no idea how that simplifys to achieve the 2 other unknown co ordinates (2,18) (-2,-2)

    I think this is totally wrong and I'm not sure what the next step is, any help would be great!
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    For question 2, you made a few mistakes with the arithmetic, but you're not really using the best method. To solve x^2(x-4)=x(4-x), note that (4-x)=-(x-4) so you have x^2(x-4)=-x(x-4). Now bring everything on one side and factorise. In this question you don't need to multiply out the brackets first.

    x^2(x-4)+x(x-4)=0

    Taking out a factor of x(x-4),

    x(x-4)[x+1]=0

    which you can solve to get x=0, 4, -1 then you can find the appropriate y coordinates.
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    For question 3b, you also made a mistake since you shouldn't be seeing x^4 when the equations you're dealing with a quadratics and cubics. Use the method of moving all the terms to one side, take out a factor of x, then you can multiply out (1+x)^2 and the whole thin should simplify down to something that you can factorise using difference of two squares.
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    a) there are only two intersections, since the curve just touches the x axis and goes back down and up through (4,0)
    b). Solve the two simultaneously. First expand the brackets. The next step is to make it equal to zero, and finally you need to factorise.The equation equal to zero should be x^3-3x^2-4x. So x(x^2-3x-4). Factorise the big bracket now. So x(x+1)(x-4)=0 so x=0 (at (0,0)), 4 and -1. Substitute these values in to find the y co-ordinates.
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    (Original post by BrainFactory)
    The steps involved: first solve the equations simultaneously to find the the values of x. So x^2(x-4)=x(4-x). Next, expand the brackets. So x^3-4x^2=4x-x^2 Next, move everything to one side so it equals zero. So x^3-4x-3x^2=0. Next, factorise this. So x(x^2-3x-4)=0 Factorise the big bracket. So x(x+1)(x-4)=0. so x= 0, 4  and  -1. Substitute these values in to either equation to find the points of intersection.
    Thank you! I understand how it work, since I already have a 2 out of 3 x's "0" and "4" this leaves me with -1 so when I put it into say "x^2(x-4) I get -1^2(-1-4)=-5, therefore my last coordinate is (-1,-5)
    thanks again,+ thumbs up post button

    (Original post by ttoby)
    For question 2, you made a few mistakes with the arithmetic, but you're not really using the best method. To solve x^2(x-4)=x(4-x), note that (4-x)=-(x-4) so you have x^2(x-4)=-x(x-4). Now bring everything on one side and factorise. In this question you don't need to multiply out the brackets first.

    x^2(x-4)+x(x-4)=0

    Taking out a factor of x(x-4),

    x(x-4)[x+1]=0

    which you can solve to get x=0, 4, -1 then you can find the appropriate y coordinates.
    Likewise, thank you for the help and the different method to solve this, although I'm a little bit stuck on the "taking out a factor of x(x-4)" I understand how that works to work out the coords of the intersection but just don't understand how to take the factor out

    (Original post by ttoby)
    For question 3b, you also made a mistake since you shouldn't be seeing x^4 when the equations you're dealing with a quadratics and cubics. Use the method of moving all the terms to one side, take out a factor of x, then you can multiply out (1+x)^2 and the whole thin should simplify down to something that you can factorise using difference of two squares.
    Yeah,I have no idea how I keep on getting x^4 I think i'm doing something wrong, I think this arithmetic is now correct but i'm not 100% as my brain has gone all numb

    2x^2+5=x+x^2-----> simplifies to-->2x^2=-4x+x^2--->0=x^2+4x

    thats where I got up to but is this correct and then would I just use d.o.t.s??

    Again, thank you for your assistance, this is really helping me out.
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    Yeah,I have no idea how I keep on getting x^4 I think i'm doing something wrong, I think this arithmetic is now correct but i'm not 100% as my brain has gone all numb

    2x^2+5=x+x^2-----> simplifies to-->2x^2=-4x+x^2--->0=x^2+4x

    thats where I got up to but is this correct and then would I just use d.o.t.s??

    Again, thank you for your assistance, this is really helping me out.
    remember that it is x(1+x)^2. This means that it is x(1+2x+x^2), which is the same as x(1+x)(1+x)=x^3+2x^2+x. To find the points of intersection, you would solve them simultaneously. So, again, x^3+2x^2+x=2x^2+5x will give you the x co-ordinates and you'll have to substitute them in again to find y.

    EDIT: So you will get x^3-4x which factorises. So x(x^2-4)=0. This gives x(x+2)(x-2)=0 so x=2, -2, 0. For practice, you should do another question - feel free to tell me what it is and what answers you got.
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    (Original post by BrainFactory)
    remember that it is x(1+x)^2. This means that it is x(1+2x+x^2), which is the same as x(1+x)(1+x)=x^3+2x^2+x. To find the points of intersection, you would solve them simultaneously. So, again, x^3+2x^2+x=2x^2+5x will give you the x co-ordinates and you'll have to substitute them in again to find y.

    EDIT: So you will get x^3-4x which factorises. So x(x^2-4)=0. This gives x(x+2)(x-2)=0 so x=2, -2, 0. For practice, you should do another question - feel free to tell me what it is and what answers you got.
    Yes thank you, you've made it very clear and consise and easy to follow, yeah I substituted the values back into the equation x(2x+5) and I got when I substituted the value 2 in as (2,18) and then the other value -2 which gave me co ordinates of (-2,-2) and obviously (0,0) and I've just checked the answers in the book and it's correct, if I could I would rep you again, but I have to spread again first :3
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    (Original post by d_aan)
    Thank you! I understand how it work, since I already have a 2 out of 3 x's "0" and "4" this leaves me with -1 so when I put it into say "x^2(x-4) I get -1^2(-1-4)=-5, therefore my last coordinate is (-1,-5)
    thanks again,+ thumbs up post button



    Likewise, thank you for the help and the different method to solve this, although I'm a little bit stuck on the "taking out a factor of x(x-4)" I understand how that works to work out the coords of the intersection but just don't understand how to take the factor out
    When you factorise something, you would have something in the form ab+ac=a(b+c). In this case, x(x-4) is your 'a', x is your b and 1 is your c. So you could think of it as x^2(x-4)+x(x-4)=x(x-4)[x]+x(x-4)[1]=x(x-4)[x+1]

    Yeah,I have no idea how I keep on getting x^4 I think i'm doing something wrong, I think this arithmetic is now correct but i'm not 100% as my brain has gone all numb

    2x^2+5=x+x^2-----> simplifies to-->2x^2=-4x+x^2--->0=x^2+4x

    thats where I got up to but is this correct and then would I just use d.o.t.s??

    Again, thank you for your assistance, this is really helping me out.
    You're still making mistakes. All I can suggest is that you leave the question for a few hours, give your brain some rest, maybe have something to eat then start the question again from scratch. I personally find that if I'm not feeling 100% then I tend to make more mistakes and I find the maths harder to understand.
 
 
 
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