Find the Maximum/Minimum Value y=x^2+8x+13

$y = x^{2} + 8 x + 13$

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 $x = - \frac{b}{2 a}$

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 $1$ .

$x = - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$x = - 4$

Evaluate $f (- 4)$ .

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

$f (- 4) = {(- 4)}^{2} + 8 (- 4) + 13$

Simplify the result.

Simplify each term.

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

$f (- 4) = 16 + 8 (- 4) + 13$

Multiply $8$ by $- 4$ .

$f (- 4) = 16 - 32 + 13$

Simplify by adding and subtracting.

Subtract $32$ from $16$ .

$f (- 4) = - 16 + 13$

Add $- 16$ and $13$ .

$f (- 4) = - 3$

The final answer is $- 3$ .

$- 3$

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

$(- 4, - 3)$

Find the Maximum/Minimum Value y=x^2+8x+13

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