Find the Maximum/Minimum Value -2x^2-6x+9

$- 2 x^{2} - 6 x + 9$

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 $b$ .

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

Remove parentheses.

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

Simplify $- \frac{- 6}{2 (- 2)}$ .

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot - 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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.

$f (- \frac{3}{2}) = - 2 {(- \frac{3}{2})}^{2} - 6 (- \frac{3}{2}) + 9$

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 $- \frac{3}{2}$ .

$f (- \frac{3}{2}) = - 2 ({(- 1)}^{2} {(\frac{3}{2})}^{2}) - 6 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 2 ({(- 1)}^{2} (\frac{3^{2}}{2^{2}})) - 6 (- \frac{3}{2}) + 9$

Raise $- 1$ to the power of $2$ .

$f (- \frac{3}{2}) = - 2 (1 (\frac{3^{2}}{2^{2}})) - 6 (- \frac{3}{2}) + 9$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$f (- \frac{3}{2}) = - 2 \frac{3^{2}}{2^{2}} - 6 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 2 \frac{9}{2^{2}} - 6 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 2 (\frac{9}{4}) - 6 (- \frac{3}{2}) + 9$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (- \frac{3}{2}) = 2 (- 1) (\frac{9}{4}) - 6 (- \frac{3}{2}) + 9$

Factor $2$ out of $4$ .

$f (- \frac{3}{2}) = 2 \cdot (- 1 \frac{9}{2 \cdot 2}) - 6 (- \frac{3}{2}) + 9$

Cancel the common factor.

$f (- \frac{3}{2}) = 2 \cdot (- 1 \frac{9}{2 \cdot 2}) - 6 (- \frac{3}{2}) + 9$

Rewrite the expression.

$f (- \frac{3}{2}) = - 1 (\frac{9}{2}) - 6 (- \frac{3}{2}) + 9$

Rewrite $- 1 (\frac{9}{2})$ as $- (\frac{9}{2})$ .

$f (- \frac{3}{2}) = - (\frac{9}{2}) - 6 (- \frac{3}{2}) + 9$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$f (- \frac{3}{2}) = - \frac{9}{2} - 6 (\frac{- 3}{2}) + 9$

Factor $2$ out of $- 6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 3$ by $- 3$ .

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

Find the common denominator.

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

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

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

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

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

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

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

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

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

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

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify each term.

Multiply $9$ by $2$ .

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

Multiply $9$ by $2$ .

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

Simplify by adding numbers.

Add $- 9$ and $18$ .

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

Add $9$ and $18$ .

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

The final answer is $\frac{27}{2}$ .

$\frac{27}{2}$

Step 4

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

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

Step 5

Find the Maximum/Minimum Value -2x^2-6x+9

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