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

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

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 $b$ .

$x = - \frac{- 8}{2 (1)}$

Remove parentheses.

$x = - \frac{- 8}{2 (1)}$

Simplify $- \frac{- 8}{2 (1)}$ .

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

Factor $2$ out of $- 8$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 4$ by $1$ .

$x = - - 4$

Multiply $- 1$ by $- 4$ .

$x = 4$

Step 3

Evaluate $f (4)$ .

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

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

Simplify the result.

Simplify each term.

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

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

Multiply $- 8$ by $4$ .

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

Simplify by subtracting numbers.

Subtract $32$ from $16$ .

$f (4) = - 16 - 16$

Subtract $16$ from $- 16$ .

$f (4) = - 32$

The final answer is $- 32$ .

$- 32$

Step 4

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

$(4, - 32)$

Step 5

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

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