Precalculus Examples

$f (x) = 4 + 3 x - x^{2}$

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 $b$ .

$x = - \frac{3}{2 (- 1)}$

Step 2.2

Remove parentheses.

$x = - \frac{3}{2 (- 1)}$

Step 2.3

Simplify $- \frac{3}{2 (- 1)}$ .

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Step 2.3.1

Multiply $2$ by $- 1$ .

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

Step 2.3.2

Move the negative in front of the fraction.

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

Step 2.3.3

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

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Step 2.3.3.1

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{3}{2})$

Step 2.3.3.2

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

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

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.

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

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

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

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Multiply $3$ by $3$ .

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.1.4

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

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

Step 3.2.2

Find the common denominator.

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Step 3.2.2.1

Write $4$ as a fraction with denominator $1$ .

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

Step 3.2.2.2

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

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

Step 3.2.2.3

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

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

Step 3.2.2.4

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

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

Step 3.2.2.5

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

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

Step 3.2.2.6

Multiply $2$ by $2$ .

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

Step 3.2.3

Combine the numerators over the common denominator.

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

Step 3.2.4

Simplify each term.

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Step 3.2.4.1

Multiply $4$ by $4$ .

$f (\frac{3}{2}) = \frac{16 + 9 \cdot 2 - 9}{4}$

Step 3.2.4.2

Multiply $9$ by $2$ .

$f (\frac{3}{2}) = \frac{16 + 18 - 9}{4}$

Step 3.2.5

Simplify by adding and subtracting.

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Step 3.2.5.1

Add $16$ and $18$ .

$f (\frac{3}{2}) = \frac{34 - 9}{4}$

Step 3.2.5.2

Subtract $9$ from $34$ .

$f (\frac{3}{2}) = \frac{25}{4}$

Step 3.2.6

The final answer is $\frac{25}{4}$ .

$\frac{25}{4}$

Step 4

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

$(\frac{3}{2}, \frac{25}{4})$

Step 5