Algebra Examples

$f (x) = - 3 x^{2} + 2 x - 5$

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 $b$ .

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

Step 2.2

Remove parentheses.

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

Step 2.3

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

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Step 2.3.1

Cancel the common factor of $2$ .

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Step 2.3.1.1

Cancel the common factor.

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

Step 2.3.1.2

Rewrite the expression.

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

Step 2.3.2

Move the negative in front of the fraction.

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

Step 2.3.3

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

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Step 2.3.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.3.2

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

$x = \frac{1}{3}$

Step 3

Evaluate $f (\frac{1}{3})$ .

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Step 3.1

Replace the variable $x$ with $\frac{1}{3}$ in the expression.

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

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

Apply the product rule to $\frac{1}{3}$ .

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

Step 3.2.1.2

One to any power is one.

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

Step 3.2.1.3

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

$f (\frac{1}{3}) = - 3 (\frac{1}{9}) + 2 (\frac{1}{3}) - 5$

Step 3.2.1.4

Cancel the common factor of $3$ .

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Step 3.2.1.4.1

Factor $3$ out of $- 3$ .

$f (\frac{1}{3}) = 3 (- 1) (\frac{1}{9}) + 2 (\frac{1}{3}) - 5$

Step 3.2.1.4.2

Factor $3$ out of $9$ .

$f (\frac{1}{3}) = 3 \cdot (- 1 \frac{1}{3 \cdot 3}) + 2 (\frac{1}{3}) - 5$

Step 3.2.1.4.3

Cancel the common factor.

$f (\frac{1}{3}) = 3 \cdot (- 1 \frac{1}{3 \cdot 3}) + 2 (\frac{1}{3}) - 5$

Step 3.2.1.4.4

Rewrite the expression.

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

Step 3.2.1.5

Rewrite $- 1 (\frac{1}{3})$ as $- (\frac{1}{3})$ .

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

Step 3.2.1.6

Combine $2$ and $\frac{1}{3}$ .

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

Step 3.2.2

Combine fractions.

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Step 3.2.2.1

Combine the numerators over the common denominator.

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

Step 3.2.2.2

Add $- 1$ and $2$ .

$f (\frac{1}{3}) = - 5 + \frac{1}{3}$

Step 3.2.3

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{1}{3}) = - 5 \cdot \frac{3}{3} + \frac{1}{3}$

Step 3.2.4

Combine $- 5$ and $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{- 5 \cdot 3}{3} + \frac{1}{3}$

Step 3.2.5

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = \frac{- 5 \cdot 3 + 1}{3}$

Step 3.2.6

Simplify the numerator.

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Step 3.2.6.1

Multiply $- 5$ by $3$ .

$f (\frac{1}{3}) = \frac{- 15 + 1}{3}$

Step 3.2.6.2

Add $- 15$ and $1$ .

$f (\frac{1}{3}) = \frac{- 14}{3}$

Step 3.2.7

Move the negative in front of the fraction.

$f (\frac{1}{3}) = - \frac{14}{3}$

Step 3.2.8

The final answer is $- \frac{14}{3}$ .

$- \frac{14}{3}$

Step 4

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

$(\frac{1}{3}, - \frac{14}{3})$

Step 5