Find the Maximum/Minimum Value -8x^2+2x-1

$- 8 x^{2} + 2 x - 1$

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 $b$ .

$x = - \frac{2}{2 (- 8)}$

Remove parentheses.

$x = - \frac{2}{2 (- 8)}$

Simplify $- \frac{2}{2 (- 8)}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$x = - \frac{2}{2 \cdot - 8}$

Rewrite the expression.

$x = - \frac{1}{- 8}$

Move the negative in front of the fraction.

$x = - - \frac{1}{8}$

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

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{1}{8})$

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

$x = \frac{1}{8}$

Step 3

Evaluate $f (\frac{1}{8})$ .

Replace the variable $x$ with $\frac{1}{8}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{8}$ .

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

One to any power is one.

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

Raise $8$ to the power of $2$ .

$f (\frac{1}{8}) = - 8 (\frac{1}{64}) + 2 (\frac{1}{8}) - 1$

Cancel the common factor of $8$ .

Factor $8$ out of $- 8$ .

$f (\frac{1}{8}) = 8 (- 1) (\frac{1}{64}) + 2 (\frac{1}{8}) - 1$

Factor $8$ out of $64$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Find the common denominator.

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

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

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

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

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

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

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

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

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

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

Reorder the factors of $4 \cdot 2$ .

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

Multiply $2$ by $4$ .

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

Combine the numerators over the common denominator.

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

Simplify the expression.

Multiply $- 1$ by $8$ .

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

Add $- 1$ and $2$ .

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

Subtract $8$ from $1$ .

$f (\frac{1}{8}) = \frac{- 7}{8}$

Move the negative in front of the fraction.

$f (\frac{1}{8}) = - \frac{7}{8}$

The final answer is $- \frac{7}{8}$ .

$- \frac{7}{8}$

Step 4

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

$(\frac{1}{8}, - \frac{7}{8})$

Step 5

Find the Maximum/Minimum Value -8x^2+2x-1

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