Precalculus Examples

f (x) = - x^{2} + 6 x - 10

$f (x) = - x^{2} + 6 x - 10$

Step 1

The maximum of a quadratic function occurs at

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

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

a

$a$ is negative, the maximum value of the function is

f (- \frac{b}{2 a})

$f (- \frac{b}{2 a})$ .

f_{max}

$f_{max}$

x = a x^{2} + b x + c

$x = a x^{2} + b x + c$ occurs at

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

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

Step 2

Find the value of

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

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

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Step 2.1

Substitute in the values of

a

$a$ and

b

$b$ .

x = - \frac{6}{2 (- 1)}

$x = - \frac{6}{2 (- 1)}$

Step 2.2

Remove parentheses.

x = - \frac{6}{2 (- 1)}

$x = - \frac{6}{2 (- 1)}$

Step 2.3

Simplify

- \frac{6}{2 (- 1)}

$- \frac{6}{2 (- 1)}$ .

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Step 2.3.1

Cancel the common factor of

6

$6$ and

2

$2$ .

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Step 2.3.1.1

Factor

2

$2$ out of

6

$6$ .

x = - \frac{2 \cdot 3}{2 \cdot - 1}

$x = - \frac{2 \cdot 3}{2 \cdot - 1}$

Step 2.3.1.2

Move the negative one from the denominator of

\frac{3}{- 1}

$\frac{3}{- 1}$ .

x = - (- 1 \cdot 3)

$x = - (- 1 \cdot 3)$

x = - (- 1 \cdot 3)

$x = - (- 1 \cdot 3)$

Step 2.3.2

Multiply

- (- 1 \cdot 3)

$- (- 1 \cdot 3)$ .

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Step 2.3.2.1

Multiply

- 1

$- 1$ by

3

$3$ .

x = - - 3

$x = - - 3$

Step 2.3.2.2

Multiply

- 1

$- 1$ by

- 3

$- 3$ .

x = 3

$x = 3$

x = 3

$x = 3$

x = 3

$x = 3$

x = 3

$x = 3$

Step 3

Evaluate

f (3)

$f (3)$ .

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Step 3.1

Replace the variable

x

$x$ with

3

$3$ in the expression.

f (3) = - {(3)}^{2} + 6 (3) - 10

$f (3) = - {(3)}^{2} + 6 (3) - 10$

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

Raise

3

$3$ to the power of

2

$2$ .

f (3) = - 1 \cdot 9 + 6 (3) - 10

$f (3) = - 1 \cdot 9 + 6 (3) - 10$

Step 3.2.1.2

Multiply

- 1

$- 1$ by

9

$9$ .

f (3) = - 9 + 6 (3) - 10

$f (3) = - 9 + 6 (3) - 10$

Step 3.2.1.3

Multiply

6

$6$ by

3

$3$ .

f (3) = - 9 + 18 - 10

$f (3) = - 9 + 18 - 10$

f (3) = - 9 + 18 - 10

$f (3) = - 9 + 18 - 10$

Step 3.2.2

Simplify by adding and subtracting.

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Step 3.2.2.1

Add

- 9

$- 9$ and

18

$18$ .

f (3) = 9 - 10

$f (3) = 9 - 10$

Step 3.2.2.2

Subtract

10

$10$ from

9

$9$ .

f (3) = - 1

$f (3) = - 1$

f (3) = - 1

$f (3) = - 1$

Step 3.2.3

The final answer is

- 1

$- 1$ .

- 1

$- 1$

- 1

$- 1$

- 1

$- 1$

Step 4

Use the

x

$x$ and

y

$y$ values to find where the maximum occurs.

(3, - 1)

$(3, - 1)$

Step 5

[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$