Find the Maximum/Minimum Value f(x)=-8x^2-8x+3

$f (x) = - 8 x^{2} - 8 x + 3$

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 $b$ .

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

Remove parentheses.

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

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

Cancel the common factor of $- 8$ and $2$ .

Factor $2$ out of $- 8$ .

$x = - \frac{2 \cdot - 4}{2 \cdot - 8}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 8$ .

$x = - \frac{2 \cdot - 4}{2 (- 8)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 4}{2 \cdot - 8}$

Rewrite the expression.

$x = - \frac{- 4}{- 8}$

Cancel the common factor of $- 4$ and $- 8$ .

Factor $- 4$ out of $- 4$ .

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

Cancel the common factors.

Factor $- 4$ out of $- 8$ .

$x = - \frac{- 4 \cdot 1}{- 4 \cdot 2}$

Cancel the common factor.

$x = - \frac{- 4 \cdot 1}{- 4 \cdot 2}$

Rewrite the expression.

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

Step 3

Evaluate $f (- \frac{1}{2})$ .

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f (- \frac{1}{2}) = - 8 {(- \frac{1}{2})}^{2} - 8 (- \frac{1}{2}) + 3$

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 $- \frac{1}{2}$ .

$f (- \frac{1}{2}) = - 8 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) - 8 (- \frac{1}{2}) + 3$

Apply the product rule to $\frac{1}{2}$ .

$f (- \frac{1}{2}) = - 8 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}})) - 8 (- \frac{1}{2}) + 3$

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

$f (- \frac{1}{2}) = - 8 (1 (\frac{1^{2}}{2^{2}})) - 8 (- \frac{1}{2}) + 3$

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

$f (- \frac{1}{2}) = - 8 \frac{1^{2}}{2^{2}} - 8 (- \frac{1}{2}) + 3$

One to any power is one.

$f (- \frac{1}{2}) = - 8 \frac{1}{2^{2}} - 8 (- \frac{1}{2}) + 3$

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

$f (- \frac{1}{2}) = - 8 (\frac{1}{4}) - 8 (- \frac{1}{2}) + 3$

Cancel the common factor of $4$ .

Factor $4$ out of $- 8$ .

$f (- \frac{1}{2}) = 4 (- 2) (\frac{1}{4}) - 8 (- \frac{1}{2}) + 3$

Cancel the common factor.

$f (- \frac{1}{2}) = 4 \cdot (- 2 (\frac{1}{4})) - 8 (- \frac{1}{2}) + 3$

Rewrite the expression.

$f (- \frac{1}{2}) = - 2 - 8 (- \frac{1}{2}) + 3$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (- \frac{1}{2}) = - 2 - 8 (\frac{- 1}{2}) + 3$

Factor $2$ out of $- 8$ .

$f (- \frac{1}{2}) = - 2 + 2 (- 4) (\frac{- 1}{2}) + 3$

Cancel the common factor.

$f (- \frac{1}{2}) = - 2 + 2 \cdot (- 4 (\frac{- 1}{2})) + 3$

Rewrite the expression.

$f (- \frac{1}{2}) = - 2 - 4 \cdot - 1 + 3$

Multiply $- 4$ by $- 1$ .

$f (- \frac{1}{2}) = - 2 + 4 + 3$

Simplify by adding numbers.

Add $- 2$ and $4$ .

$f (- \frac{1}{2}) = 2 + 3$

Add $2$ and $3$ .

$f (- \frac{1}{2}) = 5$

The final answer is $5$ .

$5$

Step 4

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

$(- \frac{1}{2}, 5)$

Step 5

Find the Maximum/Minimum Value f(x)=-8x^2-8x+3

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