Find the Maximum/Minimum Value y=2x^2+16x-8

$y = 2 x^{2} + 16 x - 8$

Step 1

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

$f_{min}$ $x = a x^{2} + b x + c$ 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 $b$ .

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

Remove parentheses.

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

Simplify $- \frac{16}{2 (2)}$ .

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

Factor $2$ out of $16$ .

$x = - \frac{2 \cdot 8}{2 \cdot 2}$

Cancel the common factors.

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

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

Cancel the common factor.

$x = - \frac{2 \cdot 8}{2 \cdot 2}$

Rewrite the expression.

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

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

$x = - \frac{2 \cdot 4}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = - \frac{2 \cdot 4}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot 4}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{4}{1}$

Divide $4$ by $1$ .

$x = - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$x = - 4$

Step 3

Evaluate $f (- 4)$ .

Replace the variable $x$ with $- 4$ in the expression.

$f (- 4) = 2 {(- 4)}^{2} + 16 (- 4) - 8$

Simplify the result.

Simplify each term.

Raise $- 4$ to the power of $2$ .

$f (- 4) = 2 \cdot 16 + 16 (- 4) - 8$

Multiply $2$ by $16$ .

$f (- 4) = 32 + 16 (- 4) - 8$

Multiply $16$ by $- 4$ .

$f (- 4) = 32 - 64 - 8$

Simplify by subtracting numbers.

Subtract $64$ from $32$ .

$f (- 4) = - 32 - 8$

Subtract $8$ from $- 32$ .

$f (- 4) = - 40$

The final answer is $- 40$ .

$- 40$

Step 4

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

$(- 4, - 40)$

Step 5

Find the Maximum/Minimum Value y=2x^2+16x-8

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