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Size:  267.0 KBthis is what i got so far not sure if im correct, cant find this question in any past paper and i dont have the mark scheme either.
    y=x^3 [standard graph]
    y=f(x-1)+2
    f(x-1)+2=(x-1)^3 +2
    ...
    y=x^3-3x^2+3x+1
    a=-3 b=3 c=1
    This is what im confused about, a the standard y=x^3 graph crosses the x-axis at -2 and 2 but this one crosses at -1 and 2 i don't know how to do this. If someone could explain how i do this it would be great.

    Thanks in advance.
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    (Original post by KappaRoss)
    ...
    Method 1:

    You need only plug the points given into the equation:0 = (-1)^3 + a(-1)^2 + b(-1) + c \iff a - b + c = 1 - this is equation 1.

    0 = (2)^3 + a(2^2) + 2b + c \iff 4a + 2b + c = -8 - this is equation 2.

    The derivative is \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 2ax + b

    You know that there is a maximum (hence zero gradient) at x=0 - so 0 = 3(0^2) + 2a(0) + b \iff b = 0

    You know that c=4 so you then have a - 0 + 4 = 1 or 4a + 0 + 4 = -8 .

    You can use either of those two equations, they'd get you the same answer.

    Method 2:

    The best way to do this question would have been to notice that x=-1 is a root, x=2 is a double root (i.e: it doesn't cut the x-axis, it just kinda touches it and bounces off tangentially).

    the equation of the cubic is y = (x+1)(x-2)^2 then expand that and find the coefficients.

    Method 3:

    Alternatively, you could also have gotten a -b + c = 1 and 4a + 2b + c = -8 then since c=4, we get two equations:

    a - b = 1 -4 \iff a- b= -3 - equation 1.

    4a + 2b = -8 - 4 = -12 \iff 2a + b = -6 - equation 2.

    Then solve simultaneously.
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    thank you very much
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    (Original post by KappaRoss)
    thank you very much
    You're very welcome!
 
 
 
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