Finite Math Examples

$f (x) = 5 x^{2} + 3 x - 7$

Find the x-value the minimum occurs at. Use this value to find the minimum value.

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The maximum or minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $(x) = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Find the value of $x$ equal to $- \frac{b}{2 a}$ .

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

Substitute in the values of $a$ and $b$ .

$x = - \frac{3}{2 (5)}$

Remove the extra parentheses from the expression $- \frac{3}{2 (5)}$ .

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Remove the parentheses from the numerator.

$x = - \frac{3}{2 (5)}$

Remove the parentheses from the denominator.

$x = - \frac{3}{2 (5)}$

Multiply $2$ by $5$ to get $2 \cdot 5$ .

$x = - \frac{3}{2 \cdot 5}$

Multiply $2$ by $5$ to get $10$ .

$x = - \frac{3}{10}$

Replace the variable $x$ with $- \frac{3}{10}$ in the expression.

$f (- \frac{3}{10}) = 5 {(- \frac{3}{10})}^{2} + 3 (- \frac{3}{10}) - 7$

Simplify the result.

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Simplify each term.

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Apply the product rule to $- \frac{3}{10}$ .

$f (- \frac{3}{10}) = 5 ({(- 1)}^{2} {(\frac{3}{10})}^{2}) + 3 (- \frac{3}{10}) - 7$

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

$f (- \frac{3}{10}) = 5 (1 {(\frac{3}{10})}^{2}) + 3 (- \frac{3}{10}) - 7$

Multiply ${(\frac{3}{10})}^{2}$ by $1$ to get ${(\frac{3}{10})}^{2}$ .

$f (- \frac{3}{10}) = 5 {(\frac{3}{10})}^{2} + 3 (- \frac{3}{10}) - 7$

Apply the product rule to $\frac{3}{10}$ .

$f (- \frac{3}{10}) = 5 (\frac{3^{2}}{10^{2}}) + 3 (- \frac{3}{10}) - 7$

Raise $3$ to the power of $2$ to get $9$ .

$f (- \frac{3}{10}) = 5 (\frac{9}{10^{2}}) + 3 (- \frac{3}{10}) - 7$

Raise $10$ to the power of $2$ to get $100$ .

$f (- \frac{3}{10}) = 5 (\frac{9}{100}) + 3 (- \frac{3}{10}) - 7$

Cancel the common factor of $5$ .

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Write $5$ as a fraction with denominator $1$ .

$f (- \frac{3}{10}) = \frac{5}{1} \cdot \frac{9}{100} + 3 (- \frac{3}{10}) - 7$

Factor out the greatest common factor $5$ .

$f (- \frac{3}{10}) = \frac{5 \cdot 1}{1} \cdot \frac{9}{5 \cdot 20} + 3 (- \frac{3}{10}) - 7$

Cancel the common factor.

$f (- \frac{3}{10}) = \frac{5 \cdot 1}{1} \cdot \frac{9}{5 \cdot 20} + 3 (- \frac{3}{10}) - 7$

Rewrite the expression.

$f (- \frac{3}{10}) = \frac{1}{1} \cdot \frac{9}{20} + 3 (- \frac{3}{10}) - 7$

Multiply $\frac{1}{1}$ and $\frac{9}{20}$ to get $\frac{9}{20}$ .

$f (- \frac{3}{10}) = \frac{9}{20} + 3 (- \frac{3}{10}) - 7$

Simplify $3 (- \frac{3}{10})$ .

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Multiply $- 1$ by $3$ to get $- 3$ .

$f (- \frac{3}{10}) = \frac{9}{20} - 3 (\frac{3}{10}) - 7$

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

$f (- \frac{3}{10}) = \frac{9}{20} + \frac{- 3}{1} \cdot \frac{3}{10} - 7$

Multiply $\frac{- 3}{1}$ and $\frac{3}{10}$ to get $\frac{- 3 \cdot 3}{10}$ .

$f (- \frac{3}{10}) = \frac{9}{20} + \frac{- 3 \cdot 3}{10} - 7$

Multiply $- 3$ by $3$ to get $- 9$ .

$f (- \frac{3}{10}) = \frac{9}{20} + \frac{- 9}{10} - 7$

Move the negative in front of the fraction.

$f (- \frac{3}{10}) = \frac{9}{20} - \frac{9}{10} - 7$

To write $- \frac{9}{10}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (- \frac{3}{10}) = \frac{9}{20} - \frac{9}{10} \cdot \frac{2}{2} - 7$

Write each expression with a common denominator of $20$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$f (- \frac{3}{10}) = \frac{9}{20} - \frac{9 \cdot 2}{10 \cdot 2} - 7$

Multiply $10$ by $2$ to get $20$ .

$f (- \frac{3}{10}) = \frac{9}{20} - \frac{9 \cdot 2}{20} - 7$

Combine the numerators over the common denominator.

$f (- \frac{3}{10}) = \frac{9 - (9 \cdot 2)}{20} - 7$

Multiply $9$ by $2$ to get $18$ .

$f (- \frac{3}{10}) = \frac{9 - 1 \cdot 18}{20} - 7$

Multiply $- 1$ by $18$ to get $- 18$ .

$f (- \frac{3}{10}) = \frac{9 - 18}{20} - 7$

Subtract $18$ from $9$ to get $- 9$ .

$f (- \frac{3}{10}) = \frac{- 9}{20} - 7$

Move the negative in front of the fraction.

$f (- \frac{3}{10}) = - \frac{9}{20} - 7$

To write $\frac{- 7}{1}$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$f (- \frac{3}{10}) = - \frac{9}{20} + \frac{- 7}{1} \cdot \frac{20}{20}$

Write each expression with a common denominator of $20$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$f (- \frac{3}{10}) = - \frac{9}{20} + \frac{- 7 \cdot 20}{1 \cdot 20}$

Multiply $20$ by $1$ to get $20$ .

$f (- \frac{3}{10}) = - \frac{9}{20} + \frac{- 7 \cdot 20}{20}$

Combine the numerators over the common denominator.

$f (- \frac{3}{10}) = \frac{- 1 \cdot 9 - 7 \cdot 20}{20}$

Simplify the numerator.

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Multiply $- 1$ by $9$ to get $- 9$ .

$f (- \frac{3}{10}) = \frac{- 9 - 7 \cdot 20}{20}$

Multiply $- 7$ by $20$ to get $- 140$ .

$f (- \frac{3}{10}) = \frac{- 9 - 140}{20}$

Subtract $140$ from $- 9$ to get $- 149$ .

$f (- \frac{3}{10}) = \frac{- 149}{20}$

Move the negative in front of the fraction.

$f (- \frac{3}{10}) = - \frac{149}{20}$

The final answer is $- \frac{149}{20}$ .

$- \frac{149}{20}$

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