Find the Maximum/Minimum Value f(x)=-x^2-3x-9

$f (x) = - x^{2} - 3 x - 9$

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 $b$ .

$x = - \frac{- 3}{2 (- 1)}$

Remove parentheses.

$x = - \frac{- 3}{2 (- 1)}$

Simplify $- \frac{- 3}{2 (- 1)}$ .

Multiply $2$ by $- 1$ .

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

Dividing two negative values results in a positive value.

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

Step 3

Evaluate $f (- \frac{3}{2})$ .

Replace the variable $x$ with $- \frac{3}{2}$ in the expression.

$f (- \frac{3}{2}) = - {(- \frac{3}{2})}^{2} - 3 (- \frac{3}{2}) - 9$

Simplify the result.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$f (- \frac{3}{2}) = - ({(- 1)}^{2} {(\frac{3}{2})}^{2}) - 3 (- \frac{3}{2}) - 9$

Apply the product rule to $\frac{3}{2}$ .

$f (- \frac{3}{2}) = - ({(- 1)}^{2} (\frac{3^{2}}{2^{2}})) - 3 (- \frac{3}{2}) - 9$

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Move ${(- 1)}^{2}$ .

$f (- \frac{3}{2}) = {(- 1)}^{2} \cdot (- 1 \frac{3^{2}}{2^{2}}) - 3 (- \frac{3}{2}) - 9$

Multiply ${(- 1)}^{2}$ by $- 1$ .

Raise $- 1$ to the power of $1$ .

$f (- \frac{3}{2}) = {(- 1)}^{2} \cdot ((- 1) (\frac{3^{2}}{2^{2}})) - 3 (- \frac{3}{2}) - 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- \frac{3}{2}) = {(- 1)}^{2 + 1} (\frac{3^{2}}{2^{2}}) - 3 (- \frac{3}{2}) - 9$

Add $2$ and $1$ .

$f (- \frac{3}{2}) = {(- 1)}^{3} (\frac{3^{2}}{2^{2}}) - 3 (- \frac{3}{2}) - 9$

Raise $- 1$ to the power of $3$ .

$f (- \frac{3}{2}) = - \frac{3^{2}}{2^{2}} - 3 (- \frac{3}{2}) - 9$

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

$f (- \frac{3}{2}) = - \frac{9}{2^{2}} - 3 (- \frac{3}{2}) - 9$

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

$f (- \frac{3}{2}) = - \frac{9}{4} - 3 (- \frac{3}{2}) - 9$

Multiply $- 3 (- \frac{3}{2})$ .

Multiply $- 1$ by $- 3$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + 3 (\frac{3}{2}) - 9$

Combine $3$ and $\frac{3}{2}$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{3 \cdot 3}{2} - 9$

Multiply $3$ by $3$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9}{2} - 9$

Find the common denominator.

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9}{2} \cdot \frac{2}{2} - 9$

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} - 9$

Write $- 9$ as a fraction with denominator $1$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + \frac{- 9}{1}$

Multiply $\frac{- 9}{1}$ by $\frac{4}{4}$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + \frac{- 9}{1} \cdot \frac{4}{4}$

Multiply $\frac{- 9}{1}$ by $\frac{4}{4}$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9 \cdot 2}{2 \cdot 2} + \frac{- 9 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (- \frac{3}{2}) = - \frac{9}{4} + \frac{9 \cdot 2}{4} + \frac{- 9 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (- \frac{3}{2}) = \frac{- 9 + 9 \cdot 2 - 9 \cdot 4}{4}$

Simplify each term.

Multiply $9$ by $2$ .

$f (- \frac{3}{2}) = \frac{- 9 + 18 - 9 \cdot 4}{4}$

Multiply $- 9$ by $4$ .

$f (- \frac{3}{2}) = \frac{- 9 + 18 - 36}{4}$

Simplify the expression.

Add $- 9$ and $18$ .

$f (- \frac{3}{2}) = \frac{9 - 36}{4}$

Subtract $36$ from $9$ .

$f (- \frac{3}{2}) = \frac{- 27}{4}$

Move the negative in front of the fraction.

$f (- \frac{3}{2}) = - \frac{27}{4}$

The final answer is $- \frac{27}{4}$ .

$- \frac{27}{4}$

Step 4

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

$(- \frac{3}{2}, - \frac{27}{4})$

Step 5

Find the Maximum/Minimum Value f(x)=-x^2-3x-9

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