Find the Maximum/Minimum Value -4x^2-12x+9

$- 4 x^{2} - 12 x + 9$

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 $b$ .

$x = - \frac{- 12}{2 (- 4)}$

Remove parentheses.

$x = - \frac{- 12}{2 (- 4)}$

Simplify $- \frac{- 12}{2 (- 4)}$ .

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$x = - \frac{2 \cdot - 6}{2 \cdot - 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 4$ .

$x = - \frac{2 \cdot - 6}{2 (- 4)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 6}{2 \cdot - 4}$

Rewrite the expression.

$x = - \frac{- 6}{- 4}$

Cancel the common factor of $- 6$ and $- 4$ .

Factor $- 2$ out of $- 6$ .

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

Cancel the common factors.

Factor $- 2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

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

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

$f (- \frac{3}{2}) = - 4 {(- \frac{3}{2})}^{2} - 12 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 4 ({(- 1)}^{2} {(\frac{3}{2})}^{2}) - 12 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 4 ({(- 1)}^{2} (\frac{3^{2}}{2^{2}})) - 12 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 4 (1 (\frac{3^{2}}{2^{2}})) - 12 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 4 \frac{3^{2}}{2^{2}} - 12 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 4 \frac{9}{2^{2}} - 12 (- \frac{3}{2}) + 9$

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

$f (- \frac{3}{2}) = - 4 (\frac{9}{4}) - 12 (- \frac{3}{2}) + 9$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

$f (- \frac{3}{2}) = 4 (- 1) (\frac{9}{4}) - 12 (- \frac{3}{2}) + 9$

Cancel the common factor.

$f (- \frac{3}{2}) = 4 \cdot (- 1 (\frac{9}{4})) - 12 (- \frac{3}{2}) + 9$

Rewrite the expression.

$f (- \frac{3}{2}) = - 1 \cdot 9 - 12 (- \frac{3}{2}) + 9$

Multiply $- 1$ by $9$ .

$f (- \frac{3}{2}) = - 9 - 12 (- \frac{3}{2}) + 9$

Cancel the common factor of $2$ .

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

$f (- \frac{3}{2}) = - 9 - 12 (\frac{- 3}{2}) + 9$

Factor $2$ out of $- 12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 6$ by $- 3$ .

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

Simplify by adding numbers.

Add $- 9$ and $18$ .

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

Add $9$ and $9$ .

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

The final answer is $18$ .

$18$

Step 4

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

$(- \frac{3}{2}, 18)$

Step 5

Find the Maximum/Minimum Value -4x^2-12x+9

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