# How do you find the maximum or minimum value of y=2x^2 +32x - 4?

Aug 13, 2017

Find the derivative, $\frac{\mathrm{dy}}{\mathrm{dx}}$, and solve for $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.
The $x$-coordinate of the turning point is at $x = - 8$.

#### Explanation:

The derivative of any term ${x}^{n}$ is given by $n {x}^{n - 1}$, so for this polynomial, $2 {x}^{2} + 32 x - 4$, the derivative is:

$2 \cdot 2 x + 1 \cdot 32 + 0 \cdot - 4$.
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 4 x + 32$.

To calculate the turning point(s), let this equal $0$.

$4 x + 32 = 0$
$\therefore x = - 8$.

By substituting this into the original equation, the turning point is at $\left(- 8 , - 137\right)$

Since $\frac{\mathrm{dy}}{\mathrm{dy}} > 0$ for $x = - 7$ and $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$ for $x = - 9$, this is a minimum turning point.