Find the Maximum/Minimum Value f(x)=-4x^2-4x+5

$f (x) = - 4 x^{2} - 4 x + 5$

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 $b$ .

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

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.

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

Step 3

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

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

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

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

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

Factor $2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 2$ by $- 1$ .

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

Simplify by adding numbers.

Add $- 1$ and $2$ .

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

Add $1$ and $5$ .

$f (- \frac{1}{2}) = 6$

The final answer is $6$ .

$6$

Step 4

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

$(- \frac{1}{2}, 6)$

Step 5

Find the Maximum/Minimum Value f(x)=-4x^2-4x+5

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