Find the Maximum/Minimum Value g(x)=-x^2+6x-5

$g (x) = - x^{2} + 6 x - 5$

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 $b$ .

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

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 $- (- 1 \cdot 3)$ .

Multiply $- 1$ by $3$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

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

$g (3) = - {(3)}^{2} + 6 (3) - 5$

Simplify the result.

Simplify each term.

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

$g (3) = - 1 \cdot 9 + 6 (3) - 5$

Multiply $- 1$ by $9$ .

$g (3) = - 9 + 6 (3) - 5$

Multiply $6$ by $3$ .

$g (3) = - 9 + 18 - 5$

Simplify by adding and subtracting.

Add $- 9$ and $18$ .

$g (3) = 9 - 5$

Subtract $5$ from $9$ .

$g (3) = 4$

The final answer is $4$ .

$4$

Step 4

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

$(3, 4)$

Step 5

Find the Maximum/Minimum Value g(x)=-x^2+6x-5

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