Algebra Examples

$y = - x^{2} + x + 6$

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

$b$ .

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

Step 2.2

Remove parentheses.

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

Step 2.3

Simplify

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

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Step 2.3.1

Cancel the common factor of

$1$ and

$- 1$ .

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Step 2.3.1.1

Rewrite

$1$ as

$- 1 (- 1)$ .

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

Step 2.3.1.2

Move the negative in front of the fraction.

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

Step 2.3.2

Multiply

$- - \frac{1}{2}$ .

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Step 2.3.2.1

Multiply

$- 1$ by

$- 1$ .

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

Step 2.3.2.2

Multiply

$\frac{1}{2}$ by

$1$ .

$x = \frac{1}{2}$

Step 3

Evaluate

$f (\frac{1}{2})$ .

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Step 3.1

Replace the variable

$x$ with

$\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = - {(\frac{1}{2})}^{2} + \frac{1}{2} + 6$

Step 3.2

Simplify the result.

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Step 3.2.1

Remove parentheses.

$f (\frac{1}{2}) = - {(\frac{1}{2})}^{2} + \frac{1}{2} + 6$

Step 3.2.2

Simplify each term.

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Step 3.2.2.1

Apply the product rule to

$\frac{1}{2}$ .

$f (\frac{1}{2}) = - \frac{1^{2}}{2^{2}} + \frac{1}{2} + 6$

Step 3.2.2.2

One to any power is one.

$f (\frac{1}{2}) = - \frac{1}{2^{2}} + \frac{1}{2} + 6$

Step 3.2.2.3

Raise

$2$ to the power of

$2$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{1}{2} + 6$

Step 3.2.3

Find the common denominator.

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Step 3.2.3.1

Multiply

$\frac{1}{2}$ by

$\frac{2}{2}$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{1}{2} \cdot \frac{2}{2} + 6$

Step 3.2.3.2

Multiply

$\frac{1}{2}$ by

$\frac{2}{2}$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + 6$

Step 3.2.3.3

Write

$6$ as a fraction with denominator

$1$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{6}{1}$

Step 3.2.3.4

Multiply

$\frac{6}{1}$ by

$\frac{4}{4}$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{6}{1} \cdot \frac{4}{4}$

Step 3.2.3.5

Multiply

$\frac{6}{1}$ by

$\frac{4}{4}$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{6 \cdot 4}{4}$

Step 3.2.3.6

Multiply

$2$ by

$2$ .

$f (\frac{1}{2}) = - \frac{1}{4} + \frac{2}{4} + \frac{6 \cdot 4}{4}$

Step 3.2.4

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{- 1 + 2 + 6 \cdot 4}{4}$

Step 3.2.5

Simplify the expression.

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Step 3.2.5.1

Multiply

$6$ by

$4$ .

$f (\frac{1}{2}) = \frac{- 1 + 2 + 24}{4}$

Step 3.2.5.2

Add

$- 1$ and

$2$ .

$f (\frac{1}{2}) = \frac{1 + 24}{4}$

Step 3.2.5.3

Add

$1$ and

$24$ .

$f (\frac{1}{2}) = \frac{25}{4}$

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.

$(\frac{1}{2}, \frac{25}{4})$

Step 5