Precalculus Examples

$f (x) = - \frac{4}{9} x^{2} + \frac{24}{9} x + 12$

Step 1

Simplify each term.

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Step 1.1

Combine $x^{2}$ and $\frac{4}{9}$ .

$f (x) = - \frac{x^{2} \cdot 4}{9} + \frac{24}{9} x + 12$

Step 1.2

Move $4$ to the left of $x^{2}$ .

$f (x) = - \frac{4 \cdot x^{2}}{9} + \frac{24}{9} x + 12$

Step 1.3

Cancel the common factor of $24$ and $9$ .

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Step 1.3.1

Factor $3$ out of $24$ .

$f (x) = - \frac{4 x^{2}}{9} + \frac{3 (8)}{9} x + 12$

Step 1.3.2

Cancel the common factors.

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Step 1.3.2.1

Factor $3$ out of $9$ .

$f (x) = - \frac{4 x^{2}}{9} + \frac{3 \cdot 8}{3 \cdot 3} x + 12$

Step 1.3.2.2

Cancel the common factor.

$f (x) = - \frac{4 x^{2}}{9} + \frac{3 \cdot 8}{3 \cdot 3} x + 12$

Step 1.3.2.3

Rewrite the expression.

$f (x) = - \frac{4 x^{2}}{9} + \frac{8}{3} x + 12$

Step 1.4

Combine $\frac{8}{3}$ and $x$ .

$f (x) = - \frac{4 x^{2}}{9} + \frac{8 x}{3} + 12$

Step 2

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 3

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

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Step 3.1

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

$x = - \frac{2.\overline{6}}{2 (- 0.\overline{4})}$

Step 3.2

Remove parentheses.

$x = - \frac{2.\overline{6}}{2 (- 0.\overline{4})}$

Step 3.3

Simplify $- \frac{2.\overline{6}}{2 (- 0.\overline{4})}$ .

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Step 3.3.1

Multiply $2$ by $- 0.\overline{4}$ .

$x = - \frac{2.\overline{6}}{- 0.\overline{8}}$

Step 3.3.2

Divide $2.\overline{6}$ by $- 0.\overline{8}$ .

$x = - - 3$

Step 3.3.3

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 4

Evaluate $f (3)$ .

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Step 4.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = - \frac{4 {(3)}^{2}}{9} + \frac{8 (3)}{3} + 12$

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $8$ by $3$ .

$f (3) = - \frac{4 {(3)}^{2}}{9} + \frac{24}{3} + 12$

Step 4.2.2

Simplify each term.

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Step 4.2.2.1

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

$f (3) = - \frac{4 \cdot 9}{9} + \frac{24}{3} + 12$

Step 4.2.2.2

Multiply $4$ by $9$ .

$f (3) = - \frac{36}{9} + \frac{24}{3} + 12$

Step 4.2.2.3

Divide $36$ by $9$ .

$f (3) = - 1 \cdot 4 + \frac{24}{3} + 12$

Step 4.2.2.4

Multiply $- 1$ by $4$ .

$f (3) = - 4 + \frac{24}{3} + 12$

Step 4.2.2.5

Divide $24$ by $3$ .

$f (3) = - 4 + 8 + 12$

Step 4.2.3

Simplify by adding numbers.

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Step 4.2.3.1

Add $- 4$ and $8$ .

$f (3) = 4 + 12$

Step 4.2.3.2

Add $4$ and $12$ .

$f (3) = 16$

Step 4.2.4

The final answer is $16$ .

$16$

Step 5

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

$(3, 16)$

Step 6