# Thread: turning point of a curve

1. ## turning 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

2. If (1, 4) is on the curve $y = ax^3 + bx^2 - 6x - 1/2$, plug in x and y to get
\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:
\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
\begin{aligned}
a + b &= 21/2 \\
3a + 2b &= 6
\end{aligned}

Solve for a and b.