Find the Maximum/Minimum Value 4x^2+4x-8

$4 x^{2} + 4 x - 8$

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 $b$ .

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

Remove parentheses.

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

Simplify $- \frac{4}{2 (4)}$ .

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$x = - \frac{2 (1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

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

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

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $- 1$ .

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

Simplify by subtracting numbers.

Subtract $2$ from $1$ .

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

Subtract $8$ from $- 1$ .

$f (- \frac{1}{2}) = - 9$

The final answer is $- 9$ .

$- 9$

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

$(- \frac{1}{2}, - 9)$

Find the Maximum/Minimum Value 4x^2+4x-8

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