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    Part a).

    So I concluded that c = 4. Then I tried to do something with like with a parabola - I took the x values to form (x+1)(x-2), but I'm not sure if I'm going the right direction on what to do really.

    Any help pls?
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    (Original post by frostyy)
    So I concluded that c = 4. Then I tried to do something with like with a parabola - I took the x values to form (x+1)(x-2), but I'm not sure if I'm going the right direction on what to do really.

    Any help pls?
    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.
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    (Original post by frostyy)
    ...
    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).

    Hence the equation of the cubic is y = (x+1)(x-2)^2 then expand that and find the coefficients.
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    (Original post by frostyy)
    ...
    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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    This question is basically seeing if you are aware of how the equation of a function relates to its roots.
    A polynomial function with roots  \alpha, \beta, \gamma, \ldots can be written as  (x-\alpha)(x-\beta)(x-\gamma) \ldots .
    A multiple root shows that at that point, the x axis is a tangent to the curve.
    A single root shows that the curve crosses the x axis at that point.
 
 
 
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