Given ( 1 , 4 ) is a turning point of the curve y = ax^3 + bx^2 - 6x - 1/2 .

Find the values of a and b

(Wondering)

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- Jul 17th 2010, 05:57 AMZharif93turning point of a curve
Given ( 1 , 4 ) is a turning point of the curve y = ax^3 + bx^2 - 6x - 1/2 .

Find the values of a and b

(Wondering) - Jul 17th 2010, 06:16 AMeumyang
If (1, 4) is on the curve $\displaystyle y = ax^3 + bx^2 - 6x - 1/2$, plug in x and y to get

$\displaystyle \begin{aligned}

4 &= a(1)^3 + b(1)^2 - 6(1) - 1/2 \\

4 &= a + b - 13/2 \\

a + b &= 21/2 \\

\end{aligned}$

If (1, 4) is a turning point, then the derivative of y is 0 at that point:

$\displaystyle \begin{aligned}

y' &= 3ax^2 + 2bx - 6 \\

0 &= 3a(1)^2 + 2b(1) - 6 \\

0 &= 3a + 2b - 6 \\

3a + 2b &= 6

\end{aligned}$

You're left with a system of equations

$\displaystyle \begin{aligned}

a + b &= 21/2 \\

3a + 2b &= 6

\end{aligned}$

Solve for a and b.