Find the Maximum/Minimum Value y=-2x^2+32x-12

$y = - 2 x^{2} + 32 x - 12$

Step 1

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{32}{2 (- 2)}$

Remove parentheses.

$x = - \frac{32}{2 (- 2)}$

Simplify $- \frac{32}{2 (- 2)}$ .

Cancel the common factor of $32$ and $2$ .

Factor $2$ out of $32$ .

$x = - \frac{2 \cdot 16}{2 \cdot - 2}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 2$ .

$x = - \frac{2 \cdot 16}{2 (- 2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 16}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{16}{- 2}$

Cancel the common factor of $16$ and $- 2$ .

Factor $2$ out of $16$ .

$x = - \frac{2 (8)}{- 2}$

Move the negative one from the denominator of $\frac{8}{- 1}$ .

$x = - (- 1 \cdot 8)$

Multiply $- (- 1 \cdot 8)$ .

Multiply $- 1$ by $8$ .

$x = - - 8$

Multiply $- 1$ by $- 8$ .

$x = 8$

Step 3

Evaluate $f (8)$ .

Replace the variable $x$ with $8$ in the expression.

$f (8) = - 2 {(8)}^{2} + 32 (8) - 12$

Simplify the result.

Simplify each term.

Raise $8$ to the power of $2$ .

$f (8) = - 2 \cdot 64 + 32 (8) - 12$

Multiply $- 2$ by $64$ .

$f (8) = - 128 + 32 (8) - 12$

Multiply $32$ by $8$ .

$f (8) = - 128 + 256 - 12$

Simplify by adding and subtracting.

Add $- 128$ and $256$ .

$f (8) = 128 - 12$

Subtract $12$ from $128$ .

$f (8) = 116$

The final answer is $116$ .

$116$

Step 4

Use the $x$ and $y$ values to find where the maximum occurs.

$(8, 116)$

Step 5

Find the Maximum/Minimum Value y=-2x^2+32x-12

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