Find the Maximum/Minimum Value -3x^2-15x+9

$- 3 x^{2} - 15 x + 9$

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 $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{- 15}{2 (- 3)}$

Remove parentheses.

$x = - \frac{- 15}{2 (- 3)}$

Cancel the common factor of $- 15$ and $- 3$ .

Factor $- 3$ out of $- 15$ .

$x = - \frac{- 3 (5)}{2 (- 3)}$

Cancel the common factors.

Factor $- 3$ out of $2 (- 3)$ .

$x = - \frac{- 3 (5)}{- 3 \cdot 2}$

Cancel the common factor.

$x = - \frac{- 3 \cdot 5}{- 3 \cdot 2}$

Rewrite the expression.

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

Evaluate $f (- \frac{5}{2})$ .

Replace the variable $x$ with $- \frac{5}{2}$ in the expression.

$f (- \frac{5}{2}) = - 3 {(- \frac{5}{2})}^{2} - 15 (- \frac{5}{2}) + 9$

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 $- \frac{5}{2}$ .

$f (- \frac{5}{2}) = - 3 ({(- 1)}^{2} {(\frac{5}{2})}^{2}) - 15 (- \frac{5}{2}) + 9$

Apply the product rule to $\frac{5}{2}$ .

$f (- \frac{5}{2}) = - 3 ({(- 1)}^{2} (\frac{5^{2}}{2^{2}})) - 15 (- \frac{5}{2}) + 9$

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

$f (- \frac{5}{2}) = - 3 (1 (\frac{5^{2}}{2^{2}})) - 15 (- \frac{5}{2}) + 9$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$f (- \frac{5}{2}) = - 3 \frac{5^{2}}{2^{2}} - 15 (- \frac{5}{2}) + 9$

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

$f (- \frac{5}{2}) = - 3 \frac{25}{2^{2}} - 15 (- \frac{5}{2}) + 9$

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

$f (- \frac{5}{2}) = - 3 (\frac{25}{4}) - 15 (- \frac{5}{2}) + 9$

Multiply $- 3 (\frac{25}{4})$ .

Combine $- 3$ and $\frac{25}{4}$ .

$f (- \frac{5}{2}) = \frac{- 3 \cdot 25}{4} - 15 (- \frac{5}{2}) + 9$

Multiply $- 3$ by $25$ .

$f (- \frac{5}{2}) = \frac{- 75}{4} - 15 (- \frac{5}{2}) + 9$

Move the negative in front of the fraction.

$f (- \frac{5}{2}) = - \frac{75}{4} - 15 (- \frac{5}{2}) + 9$

Multiply $- 15 (- \frac{5}{2})$ .

Multiply $- 1$ by $- 15$ .

$f (- \frac{5}{2}) = - \frac{75}{4} + 15 (\frac{5}{2}) + 9$

Combine $15$ and $\frac{5}{2}$ .

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{15 \cdot 5}{2} + 9$

Multiply $15$ by $5$ .

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75}{2} + 9$

Find the common denominator.

Multiply $\frac{75}{2}$ by $\frac{2}{2}$ .

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75}{2} \cdot \frac{2}{2} + 9$

Multiply $\frac{75}{2}$ by $\frac{2}{2}$ .

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75 \cdot 2}{2 \cdot 2} + 9$

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

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75 \cdot 2}{2 \cdot 2} + \frac{9}{1}$

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

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75 \cdot 2}{2 \cdot 2} + \frac{9}{1} \cdot \frac{4}{4}$

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

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75 \cdot 2}{2 \cdot 2} + \frac{9 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (- \frac{5}{2}) = - \frac{75}{4} + \frac{75 \cdot 2}{4} + \frac{9 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (- \frac{5}{2}) = \frac{- 75 + 75 \cdot 2 + 9 \cdot 4}{4}$

Simplify each term.

Multiply $75$ by $2$ .

$f (- \frac{5}{2}) = \frac{- 75 + 150 + 9 \cdot 4}{4}$

Multiply $9$ by $4$ .

$f (- \frac{5}{2}) = \frac{- 75 + 150 + 36}{4}$

Simplify by adding numbers.

Add $- 75$ and $150$ .

$f (- \frac{5}{2}) = \frac{75 + 36}{4}$

Add $75$ and $36$ .

$f (- \frac{5}{2}) = \frac{111}{4}$

The final answer is $\frac{111}{4}$ .

$\frac{111}{4}$

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

$(- \frac{5}{2}, \frac{111}{4})$

Find the Maximum/Minimum Value -3x^2-15x+9

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