Find the Maximum/Minimum Value y=-2x^2-x+15

$y = - 2 x^{2} - x + 15$

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

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

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

Remove parentheses.

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

Simplify $- \frac{- 1}{2 (- 2)}$ .

Multiply $2$ by $- 2$ .

$x = - \frac{- 1}{- 4}$

Dividing two negative values results in a positive value.

$x = - \frac{1}{4}$

Step 3

Evaluate $f (- \frac{1}{4})$ .

Replace the variable $x$ with $- \frac{1}{4}$ in the expression.

$f (- \frac{1}{4}) = - 2 {(- \frac{1}{4})}^{2} - (- \frac{1}{4}) + 15$

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 $- \frac{1}{4}$ .

$f (- \frac{1}{4}) = - 2 ({(- 1)}^{2} {(\frac{1}{4})}^{2}) - (- \frac{1}{4}) + 15$

Apply the product rule to $\frac{1}{4}$ .

$f (- \frac{1}{4}) = - 2 ({(- 1)}^{2} (\frac{1^{2}}{4^{2}})) - (- \frac{1}{4}) + 15$

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

$f (- \frac{1}{4}) = - 2 (1 (\frac{1^{2}}{4^{2}})) - (- \frac{1}{4}) + 15$

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

$f (- \frac{1}{4}) = - 2 \frac{1^{2}}{4^{2}} - (- \frac{1}{4}) + 15$

One to any power is one.

$f (- \frac{1}{4}) = - 2 \frac{1}{4^{2}} - (- \frac{1}{4}) + 15$

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

$f (- \frac{1}{4}) = - 2 (\frac{1}{16}) - (- \frac{1}{4}) + 15$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (- \frac{1}{4}) = 2 (- 1) (\frac{1}{16}) - (- \frac{1}{4}) + 15$

Factor $2$ out of $16$ .

$f (- \frac{1}{4}) = 2 \cdot (- 1 \frac{1}{2 \cdot 8}) - (- \frac{1}{4}) + 15$

Cancel the common factor.

$f (- \frac{1}{4}) = 2 \cdot (- 1 \frac{1}{2 \cdot 8}) - (- \frac{1}{4}) + 15$

Rewrite the expression.

$f (- \frac{1}{4}) = - 1 (\frac{1}{8}) - (- \frac{1}{4}) + 15$

Rewrite $- 1 (\frac{1}{8})$ as $- (\frac{1}{8})$ .

$f (- \frac{1}{4}) = - (\frac{1}{8}) - (- \frac{1}{4}) + 15$

Multiply $- (- \frac{1}{4})$ .

Multiply $- 1$ by $- 1$ .

$f (- \frac{1}{4}) = - \frac{1}{8} + 1 (\frac{1}{4}) + 15$

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

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{1}{4} + 15$

Find the common denominator.

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

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{1}{4} \cdot \frac{2}{2} + 15$

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

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{2}{4 \cdot 2} + 15$

Write $15$ as a fraction with denominator $1$ .

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{2}{4 \cdot 2} + \frac{15}{1}$

Multiply $\frac{15}{1}$ by $\frac{8}{8}$ .

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{2}{4 \cdot 2} + \frac{15}{1} \cdot \frac{8}{8}$

Multiply $\frac{15}{1}$ by $\frac{8}{8}$ .

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{2}{4 \cdot 2} + \frac{15 \cdot 8}{8}$

Reorder the factors of $4 \cdot 2$ .

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{2}{2 \cdot 4} + \frac{15 \cdot 8}{8}$

Multiply $2$ by $4$ .

$f (- \frac{1}{4}) = - \frac{1}{8} + \frac{2}{8} + \frac{15 \cdot 8}{8}$

Combine the numerators over the common denominator.

$f (- \frac{1}{4}) = \frac{- 1 + 2 + 15 \cdot 8}{8}$

Simplify the expression.

Multiply $15$ by $8$ .

$f (- \frac{1}{4}) = \frac{- 1 + 2 + 120}{8}$

Add $- 1$ and $2$ .

$f (- \frac{1}{4}) = \frac{1 + 120}{8}$

Add $1$ and $120$ .

$f (- \frac{1}{4}) = \frac{121}{8}$

The final answer is $\frac{121}{8}$ .

$\frac{121}{8}$

Step 4

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

$(- \frac{1}{4}, \frac{121}{8})$

Step 5

Find the Maximum/Minimum Value y=-2x^2-x+15

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