Find the Maximum/Minimum Value y=-x^2+8x-3

$y = - x^{2} + 8 x - 3$

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 $b$ .

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

Remove parentheses.

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

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

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

Factor $2$ out of $8$ .

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

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$x = - (- 1 \cdot 4)$

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$x = - - 4$

Multiply $- 1$ by $- 4$ .

$x = 4$

Step 3

Evaluate $f (4)$ .

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

$f (4) = - {(4)}^{2} + 8 (4) - 3$

Simplify the result.

Simplify each term.

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

$f (4) = - 1 \cdot 16 + 8 (4) - 3$

Multiply $- 1$ by $16$ .

$f (4) = - 16 + 8 (4) - 3$

Multiply $8$ by $4$ .

$f (4) = - 16 + 32 - 3$

Simplify by adding and subtracting.

Add $- 16$ and $32$ .

$f (4) = 16 - 3$

Subtract $3$ from $16$ .

$f (4) = 13$

The final answer is $13$ .

$13$

Step 4

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

$(4, 13)$

Step 5

Find the Maximum/Minimum Value y=-x^2+8x-3

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