Precalculus Examples

$f (x) = - 5 x^{2} - 15 x$

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

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Step 2.1

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

$x = - \frac{- 15}{2 (- 5)}$

Step 2.2

Remove parentheses.

$x = - \frac{- 15}{2 (- 5)}$

Step 2.3

Cancel the common factor of $- 15$ and $- 5$ .

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Step 2.3.1

Factor $- 5$ out of $- 15$ .

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

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor $- 5$ out of $2 (- 5)$ .

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

Step 2.3.2.2

Cancel the common factor.

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

Step 2.3.2.3

Rewrite the expression.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

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

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.1.4

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

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

Step 3.2.1.5

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

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

Step 3.2.1.6

Multiply $- 5 (\frac{9}{4})$ .

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Step 3.2.1.6.1

Combine $- 5$ and $\frac{9}{4}$ .

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

Step 3.2.1.6.2

Multiply $- 5$ by $9$ .

$f (- \frac{3}{2}) = \frac{- 45}{4} - 15 (- \frac{3}{2})$

Step 3.2.1.7

Move the negative in front of the fraction.

$f (- \frac{3}{2}) = - \frac{45}{4} - 15 (- \frac{3}{2})$

Step 3.2.1.8

Multiply $- 15 (- \frac{3}{2})$ .

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Step 3.2.1.8.1

Multiply $- 1$ by $- 15$ .

$f (- \frac{3}{2}) = - \frac{45}{4} + 15 (\frac{3}{2})$

Step 3.2.1.8.2

Combine $15$ and $\frac{3}{2}$ .

$f (- \frac{3}{2}) = - \frac{45}{4} + \frac{15 \cdot 3}{2}$

Step 3.2.1.8.3

Multiply $15$ by $3$ .

$f (- \frac{3}{2}) = - \frac{45}{4} + \frac{45}{2}$

Step 3.2.2

To write $\frac{45}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (- \frac{3}{2}) = - \frac{45}{4} + \frac{45}{2} \cdot \frac{2}{2}$

Step 3.2.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.2.3.1

Multiply $\frac{45}{2}$ by $\frac{2}{2}$ .

$f (- \frac{3}{2}) = - \frac{45}{4} + \frac{45 \cdot 2}{2 \cdot 2}$

Step 3.2.3.2

Multiply $2$ by $2$ .

$f (- \frac{3}{2}) = - \frac{45}{4} + \frac{45 \cdot 2}{4}$

Step 3.2.4

Combine the numerators over the common denominator.

$f (- \frac{3}{2}) = \frac{- 45 + 45 \cdot 2}{4}$

Step 3.2.5

Simplify the numerator.

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Step 3.2.5.1

Multiply $45$ by $2$ .

$f (- \frac{3}{2}) = \frac{- 45 + 90}{4}$

Step 3.2.5.2

Add $- 45$ and $90$ .

$f (- \frac{3}{2}) = \frac{45}{4}$

Step 3.2.6

The final answer is $\frac{45}{4}$ .

$\frac{45}{4}$

Step 4

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

$(- \frac{3}{2}, \frac{45}{4})$

Step 5