Find the Maximum/Minimum Value -x^2+10x-29

$- x^{2} + 10 x - 29$

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 $b$ .

$x = - \frac{10}{2 (- 1)}$

Remove parentheses.

$x = - \frac{10}{2 (- 1)}$

Simplify $- \frac{10}{2 (- 1)}$ .

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

$x = - \frac{2 \cdot 5}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$x = - (- 1 \cdot 5)$

Multiply $- (- 1 \cdot 5)$ .

Multiply $- 1$ by $5$ .

$x = - - 5$

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 3

Evaluate $f (5)$ .

Replace the variable $x$ with $5$ in the expression.

$f (5) = - {(5)}^{2} + 10 (5) - 29$

Simplify the result.

Simplify each term.

Raise $5$ to the power of $2$ .

$f (5) = - 1 \cdot 25 + 10 (5) - 29$

Multiply $- 1$ by $25$ .

$f (5) = - 25 + 10 (5) - 29$

Multiply $10$ by $5$ .

$f (5) = - 25 + 50 - 29$

Simplify by adding and subtracting.

Add $- 25$ and $50$ .

$f (5) = 25 - 29$

Subtract $29$ from $25$ .

$f (5) = - 4$

The final answer is $- 4$ .

$- 4$

Step 4

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

$(5, - 4)$

Step 5

Find the Maximum/Minimum Value -x^2+10x-29

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