Find the Maximum/Minimum Value 5x^2+13x-10

$5 x^{2} + 13 x - 10$

Step 1

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . 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 $x = - \frac{b}{2 a}$

Step 2

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

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

$x = - \frac{13}{2 (5)}$

Remove parentheses.

$x = - \frac{13}{2 (5)}$

Multiply $2$ by $5$ .

$x = - \frac{13}{10}$

Step 3

Evaluate $f (- \frac{13}{10})$ .

Replace the variable $x$ with $- \frac{13}{10}$ in the expression.

$f (- \frac{13}{10}) = 5 {(- \frac{13}{10})}^{2} + 13 (- \frac{13}{10}) - 10$

Simplify the result.

Simplify each term.

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

Apply the product rule to $- \frac{13}{10}$ .

$f (- \frac{13}{10}) = 5 ({(- 1)}^{2} {(\frac{13}{10})}^{2}) + 13 (- \frac{13}{10}) - 10$

Apply the product rule to $\frac{13}{10}$ .

$f (- \frac{13}{10}) = 5 ({(- 1)}^{2} (\frac{13^{2}}{10^{2}})) + 13 (- \frac{13}{10}) - 10$

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

$f (- \frac{13}{10}) = 5 (1 (\frac{13^{2}}{10^{2}})) + 13 (- \frac{13}{10}) - 10$

Multiply $\frac{13^{2}}{10^{2}}$ by $1$ .

$f (- \frac{13}{10}) = 5 (\frac{13^{2}}{10^{2}}) + 13 (- \frac{13}{10}) - 10$

Raise $13$ to the power of $2$ .

$f (- \frac{13}{10}) = 5 (\frac{169}{10^{2}}) + 13 (- \frac{13}{10}) - 10$

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

$f (- \frac{13}{10}) = 5 (\frac{169}{100}) + 13 (- \frac{13}{10}) - 10$

Cancel the common factor of $5$ .

Factor $5$ out of $100$ .

$f (- \frac{13}{10}) = 5 (\frac{169}{5 (20)}) + 13 (- \frac{13}{10}) - 10$

Cancel the common factor.

$f (- \frac{13}{10}) = 5 (\frac{169}{5 \cdot 20}) + 13 (- \frac{13}{10}) - 10$

Rewrite the expression.

$f (- \frac{13}{10}) = \frac{169}{20} + 13 (- \frac{13}{10}) - 10$

Multiply $13 (- \frac{13}{10})$ .

Multiply $- 1$ by $13$ .

$f (- \frac{13}{10}) = \frac{169}{20} - 13 (\frac{13}{10}) - 10$

Combine $- 13$ and $\frac{13}{10}$ .

$f (- \frac{13}{10}) = \frac{169}{20} + \frac{- 13 \cdot 13}{10} - 10$

Multiply $- 13$ by $13$ .

$f (- \frac{13}{10}) = \frac{169}{20} + \frac{- 169}{10} - 10$

Move the negative in front of the fraction.

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169}{10} - 10$

Find the common denominator.

Multiply $\frac{169}{10}$ by $\frac{2}{2}$ .

$f (- \frac{13}{10}) = \frac{169}{20} - (\frac{169}{10} \cdot \frac{2}{2}) - 10$

Multiply $\frac{169}{10}$ by $\frac{2}{2}$ .

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169 \cdot 2}{10 \cdot 2} - 10$

Write $- 10$ as a fraction with denominator $1$ .

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169 \cdot 2}{10 \cdot 2} + \frac{- 10}{1}$

Multiply $\frac{- 10}{1}$ by $\frac{20}{20}$ .

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169 \cdot 2}{10 \cdot 2} + \frac{- 10}{1} \cdot \frac{20}{20}$

Multiply $\frac{- 10}{1}$ by $\frac{20}{20}$ .

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169 \cdot 2}{10 \cdot 2} + \frac{- 10 \cdot 20}{20}$

Reorder the factors of $10 \cdot 2$ .

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169 \cdot 2}{2 \cdot 10} + \frac{- 10 \cdot 20}{20}$

Multiply $2$ by $10$ .

$f (- \frac{13}{10}) = \frac{169}{20} - \frac{169 \cdot 2}{20} + \frac{- 10 \cdot 20}{20}$

Combine the numerators over the common denominator.

$f (- \frac{13}{10}) = \frac{169 - 169 \cdot 2 - 10 \cdot 20}{20}$

Simplify each term.

Multiply $- 169$ by $2$ .

$f (- \frac{13}{10}) = \frac{169 - 338 - 10 \cdot 20}{20}$

Multiply $- 10$ by $20$ .

$f (- \frac{13}{10}) = \frac{169 - 338 - 200}{20}$

Simplify the expression.

Subtract $338$ from $169$ .

$f (- \frac{13}{10}) = \frac{- 169 - 200}{20}$

Subtract $200$ from $- 169$ .

$f (- \frac{13}{10}) = \frac{- 369}{20}$

Move the negative in front of the fraction.

$f (- \frac{13}{10}) = - \frac{369}{20}$

The final answer is $- \frac{369}{20}$ .

$- \frac{369}{20}$

Step 4

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

$(- \frac{13}{10}, - \frac{369}{20})$

Step 5

Find the Maximum/Minimum Value 5x^2+13x-10

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