Find the Maximum/Minimum Value y=-x^2+6x-8

$y = - x^{2} + 6 x - 8$

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

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

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

$x = - \frac{6}{2 (- 1)}$

Remove parentheses.

$x = - \frac{6}{2 (- 1)}$

Simplify $- \frac{6}{2 (- 1)}$ .

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$x = - (- 1 \cdot 3)$

Multiply.

Multiply $- 1$ by $3$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Evaluate $f (3)$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = - {(3)}^{2} + 6 (3) - 8$

Simplify the result.

Simplify each term.

Raise $3$ to the power of $2$ .

$f (3) = - 1 \cdot 9 + 6 (3) - 8$

Multiply $- 1$ by $9$ .

$f (3) = - 9 + 6 (3) - 8$

Multiply $6$ by $3$ .

$f (3) = - 9 + 18 - 8$

Simplify by adding and subtracting.

Add $- 9$ and $18$ .

$f (3) = 9 - 8$

Subtract $8$ from $9$ .

$f (3) = 1$

The final answer is $1$ .

$1$

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

$(3, 1)$

Find the Maximum/Minimum Value y=-x^2+6x-8

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