Algebra Examples

$- 5 x^{2} - 5 x + 3$

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 $b$ .

$x = - \frac{- 5}{2 (- 5)}$

Step 2.2

Remove parentheses.

$x = - \frac{- 5}{2 (- 5)}$

Step 2.3

Cancel the common factor of $- 5$ .

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Step 2.3.1

Cancel the common factor.

$x = - \frac{- 5}{2 \cdot - 5}$

Step 2.3.2

Rewrite the expression.

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

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.

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

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.2.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

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

Step 3.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

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

Step 3.2.1.2

Raise $- 1$ to the power of $2$ .

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

Step 3.2.1.3

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

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

Step 3.2.1.4

One to any power is one.

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

Step 3.2.1.5

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

$f (- \frac{1}{2}) = - 5 (\frac{1}{4}) - 5 (- \frac{1}{2}) + 3$

Step 3.2.1.6

Combine $- 5$ and $\frac{1}{4}$ .

$f (- \frac{1}{2}) = \frac{- 5}{4} - 5 (- \frac{1}{2}) + 3$

Step 3.2.1.7

Move the negative in front of the fraction.

$f (- \frac{1}{2}) = - \frac{5}{4} - 5 (- \frac{1}{2}) + 3$

Step 3.2.1.8

Multiply $- 5 (- \frac{1}{2})$ .

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Step 3.2.1.8.1

Multiply $- 1$ by $- 5$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + 5 (\frac{1}{2}) + 3$

Step 3.2.1.8.2

Combine $5$ and $\frac{1}{2}$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5}{2} + 3$

Step 3.2.2

Find the common denominator.

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Step 3.2.2.1

Multiply $\frac{5}{2}$ by $\frac{2}{2}$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5}{2} \cdot \frac{2}{2} + 3$

Step 3.2.2.2

Multiply $\frac{5}{2}$ by $\frac{2}{2}$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5 \cdot 2}{2 \cdot 2} + 3$

Step 3.2.2.3

Write $3$ as a fraction with denominator $1$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5 \cdot 2}{2 \cdot 2} + \frac{3}{1}$

Step 3.2.2.4

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5 \cdot 2}{2 \cdot 2} + \frac{3}{1} \cdot \frac{4}{4}$

Step 3.2.2.5

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5 \cdot 2}{2 \cdot 2} + \frac{3 \cdot 4}{4}$

Step 3.2.2.6

Multiply $2$ by $2$ .

$f (- \frac{1}{2}) = - \frac{5}{4} + \frac{5 \cdot 2}{4} + \frac{3 \cdot 4}{4}$

Step 3.2.3

Combine the numerators over the common denominator.

$f (- \frac{1}{2}) = \frac{- 5 + 5 \cdot 2 + 3 \cdot 4}{4}$

Step 3.2.4

Simplify each term.

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Step 3.2.4.1

Multiply $5$ by $2$ .

$f (- \frac{1}{2}) = \frac{- 5 + 10 + 3 \cdot 4}{4}$

Step 3.2.4.2

Multiply $3$ by $4$ .

$f (- \frac{1}{2}) = \frac{- 5 + 10 + 12}{4}$

Step 3.2.5

Simplify by adding numbers.

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Step 3.2.5.1

Add $- 5$ and $10$ .

$f (- \frac{1}{2}) = \frac{5 + 12}{4}$

Step 3.2.5.2

Add $5$ and $12$ .

$f (- \frac{1}{2}) = \frac{17}{4}$

Step 3.2.6

The final answer is $\frac{17}{4}$ .

$\frac{17}{4}$

Step 4

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

$(- \frac{1}{2}, \frac{17}{4})$

Step 5