Find the Maximum/Minimum Value f(x)=4x^2+8x-60

$f (x) = 4 x^{2} + 8 x - 60$

Step 1

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

Step 2

Find the value of $x = - \frac{b}{2 a}$ .

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

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

Remove parentheses.

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

Simplify $- \frac{8}{2 (4)}$ .

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

Factor $2$ out of $8$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 4 {(- 1)}^{2} + 8 (- 1) - 60$

Simplify the result.

Simplify each term.

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

$f (- 1) = 4 \cdot 1 + 8 (- 1) - 60$

Multiply $4$ by $1$ .

$f (- 1) = 4 + 8 (- 1) - 60$

Multiply $8$ by $- 1$ .

$f (- 1) = 4 - 8 - 60$

Simplify by subtracting numbers.

Subtract $8$ from $4$ .

$f (- 1) = - 4 - 60$

Subtract $60$ from $- 4$ .

$f (- 1) = - 64$

The final answer is $- 64$ .

$- 64$

Step 4

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

$(- 1, - 64)$

Step 5

Find the Maximum/Minimum Value f(x)=4x^2+8x-60

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