Find the Maximum/Minimum Value f(x)=2x^2+5x-6

$f (x) = 2 x^{2} + 5 x - 6$

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

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

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

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

Remove parentheses.

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

Multiply $2$ by $2$ .

$x = - \frac{5}{4}$

Evaluate $f (- \frac{5}{4})$ .

Replace the variable $x$ with $- \frac{5}{4}$ in the expression.

$f (- \frac{5}{4}) = 2 {(- \frac{5}{4})}^{2} + 5 (- \frac{5}{4}) - 6$

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 $- \frac{5}{4}$ .

$f (- \frac{5}{4}) = 2 ({(- 1)}^{2} {(\frac{5}{4})}^{2}) + 5 (- \frac{5}{4}) - 6$

Apply the product rule to $\frac{5}{4}$ .

$f (- \frac{5}{4}) = 2 ({(- 1)}^{2} (\frac{5^{2}}{4^{2}})) + 5 (- \frac{5}{4}) - 6$

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

$f (- \frac{5}{4}) = 2 (1 (\frac{5^{2}}{4^{2}})) + 5 (- \frac{5}{4}) - 6$

Multiply $\frac{5^{2}}{4^{2}}$ by $1$ .

$f (- \frac{5}{4}) = 2 (\frac{5^{2}}{4^{2}}) + 5 (- \frac{5}{4}) - 6$

Raise $5$ to the power of $2$ .

$f (- \frac{5}{4}) = 2 (\frac{25}{4^{2}}) + 5 (- \frac{5}{4}) - 6$

Raise $4$ to the power of $2$ .

$f (- \frac{5}{4}) = 2 (\frac{25}{16}) + 5 (- \frac{5}{4}) - 6$

Cancel the common factor of $2$ .

Factor $2$ out of $16$ .

$f (- \frac{5}{4}) = 2 (\frac{25}{2 (8)}) + 5 (- \frac{5}{4}) - 6$

Cancel the common factor.

$f (- \frac{5}{4}) = 2 (\frac{25}{2 \cdot 8}) + 5 (- \frac{5}{4}) - 6$

Rewrite the expression.

$f (- \frac{5}{4}) = \frac{25}{8} + 5 (- \frac{5}{4}) - 6$

Multiply $5 (- \frac{5}{4})$ .

Multiply $- 1$ by $5$ .

$f (- \frac{5}{4}) = \frac{25}{8} - 5 (\frac{5}{4}) - 6$

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

$f (- \frac{5}{4}) = \frac{25}{8} + \frac{- 5 \cdot 5}{4} - 6$

Multiply $- 5$ by $5$ .

$f (- \frac{5}{4}) = \frac{25}{8} + \frac{- 25}{4} - 6$

Move the negative in front of the fraction.

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25}{4} - 6$

Find the common denominator.

Multiply $\frac{25}{4}$ by $\frac{2}{2}$ .

$f (- \frac{5}{4}) = \frac{25}{8} - (\frac{25}{4} \cdot \frac{2}{2}) - 6$

Multiply $\frac{25}{4}$ by $\frac{2}{2}$ .

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25 \cdot 2}{4 \cdot 2} - 6$

Write $- 6$ as a fraction with denominator $1$ .

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25 \cdot 2}{4 \cdot 2} + \frac{- 6}{1}$

Multiply $\frac{- 6}{1}$ by $\frac{8}{8}$ .

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25 \cdot 2}{4 \cdot 2} + \frac{- 6}{1} \cdot \frac{8}{8}$

Multiply $\frac{- 6}{1}$ by $\frac{8}{8}$ .

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25 \cdot 2}{4 \cdot 2} + \frac{- 6 \cdot 8}{8}$

Reorder the factors of $4 \cdot 2$ .

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25 \cdot 2}{2 \cdot 4} + \frac{- 6 \cdot 8}{8}$

Multiply $2$ by $4$ .

$f (- \frac{5}{4}) = \frac{25}{8} - \frac{25 \cdot 2}{8} + \frac{- 6 \cdot 8}{8}$

Combine the numerators over the common denominator.

$f (- \frac{5}{4}) = \frac{25 - 25 \cdot 2 - 6 \cdot 8}{8}$

Simplify each term.

Multiply $- 25$ by $2$ .

$f (- \frac{5}{4}) = \frac{25 - 50 - 6 \cdot 8}{8}$

Multiply $- 6$ by $8$ .

$f (- \frac{5}{4}) = \frac{25 - 50 - 48}{8}$

Simplify the expression.

Subtract $50$ from $25$ .

$f (- \frac{5}{4}) = \frac{- 25 - 48}{8}$

Subtract $48$ from $- 25$ .

$f (- \frac{5}{4}) = \frac{- 73}{8}$

Move the negative in front of the fraction.

$f (- \frac{5}{4}) = - \frac{73}{8}$

The final answer is $- \frac{73}{8}$ .

$- \frac{73}{8}$

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

$(- \frac{5}{4}, - \frac{73}{8})$

Find the Maximum/Minimum Value f(x)=2x^2+5x-6

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