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Math Help - Cubic Functions Derivatives

  1. #1
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    Cubic Functions Derivatives

    Hello all, i need assistance with these, thanks

    1.) Find a cubic function whose graph touches the x-axis at x = -4, passes through the origin and has a value of 10 when x = 5.


    2.) The function f(x) = 2x^3 + ax^2 - bx +3. When f(x) is divided by x - 2 the remainder is 15 and f(1) = 0

    A.) Calculate the values of a and b
    B.) Find the other two factors of f(x)

    3.) The polynomial P(x) = x^3 + ax^2 + bx - 9 has zeros at x = 1 and x = -3. The values of a and b are?

    Please guys, full working out and answer, me be stupid :P
    this isnt homework, its revision so you dont have to worry about me cheating. THANKS!
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  2. #2
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    Quote Originally Posted by andrew2322 View Post
    Hello all, i need assistance with these:

    1.) Find a cubic function whose graph touches the x-axis at x = -4, passes through the origin and has a value of 10 when x = 5.
    ...
    The general equation of a cubic function is:

    f(x)=ax^3+bx^2+cx+d and it's first derivative

    f'(x)=3ax^2+2bx+c

    You already know:

    \begin{array}{l}f(0)=0 \\ f(-4)=0 \\ f'(-4)=0 \\ f(5)=10\end{array} ...... ~\Longrightarrow~...... \begin{array}{l}d=0 \\ -64a+16b-4c =0 \\ 48a - 8b + c=0 \\ 125a+25b+5c=10\end{array}

    Solve the system of simultaneous equations.
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  3. #3
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    Quote Originally Posted by andrew2322 View Post
    Hello all, i need assistance with these...


    2.) The function f(x) = 2x^3 + ax^2 - bx +3. When f(x) is divided by x - 2 the remainder is 15 and f(1) = 0

    A.) Calculate the values of a and b
    B.) Find the other two factors of f(x)

    ...
    to A.):

    From f(1) = 0 you know that

    a-b+5=0...... [1]

    Use long divison to calculate the remainder. I've got as remainder r = 19 +4a - 2b which is according to your problem 15:

    19+4a-2b = 15 ...... [2]

    Solve this system of simultaneous equations for a and b. I've got a = 3 and b = 8

    Thus the equation of the function becomes:

    f(x)=2x^2+3x^2-8x+3

    to B.):

    Use synthetic division :

    (2x^2+3x^2-8x+3) \div (x-1) = 2x^2+5x-3

    Now solve the quadratic equation

    2x^2+5x-3 = 0

    You should get x = \frac12~\vee~x=-3 and therefore the other 2 factors are:

    (x+3)\left(x-\frac12\right)
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    hey

    im so confused lol
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