Find the Maximum/Minimum Value y=-x^2-10x-30

$y = - x^{2} - 10 x - 30$

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 $\frac{- 5}{- 1}$ .

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

Rewrite $- 1 \cdot - 5$ as $- - 5$ .

$x = - - - 5$

Multiply $- - - 5$ .

Multiply $- 1$ by $- 5$ .

$x = - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$x = - 5$

Step 3

Evaluate $f (- 5)$ .

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

$f (- 5) = - {(- 5)}^{2} - 10 \cdot - 5 - 30$

Simplify the result.

Simplify each term.

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

$f (- 5) = - 1 \cdot 25 - 10 \cdot - 5 - 30$

Multiply $- 1$ by $25$ .

$f (- 5) = - 25 - 10 \cdot - 5 - 30$

Multiply $- 10$ by $- 5$ .

$f (- 5) = - 25 + 50 - 30$

Simplify by adding and subtracting.

Add $- 25$ and $50$ .

$f (- 5) = 25 - 30$

Subtract $30$ from $25$ .

$f (- 5) = - 5$

The final answer is $- 5$ .

$- 5$

Step 4

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

$(- 5, - 5)$

Step 5

Find the Maximum/Minimum Value y=-x^2-10x-30

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