Find the Maximum/Minimum Value f(x)=8x^2-64x+2

$f (x) = 8 x^{2} - 64 x + 2$

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 $b$ .

$x = - \frac{- 64}{2 (8)}$

Remove parentheses.

$x = - \frac{- 64}{2 (8)}$

Simplify $- \frac{- 64}{2 (8)}$ .

Cancel the common factor of $- 64$ and $2$ .

Factor $2$ out of $- 64$ .

$x = - \frac{2 \cdot - 32}{2 \cdot 8}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 32}{2 (8)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 32}{2 \cdot 8}$

Rewrite the expression.

$x = - \frac{- 32}{8}$

Cancel the common factor of $- 32$ and $8$ .

Factor $8$ out of $- 32$ .

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

Cancel the common factors.

Factor $8$ out of $8$ .

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

Cancel the common factor.

$x = - \frac{8 \cdot - 4}{8 \cdot 1}$

Rewrite the expression.

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

Divide $- 4$ by $1$ .

$x = - - 4$

Multiply $- 1$ by $- 4$ .

$x = 4$

Step 3

Evaluate $f (4)$ .

Replace the variable $x$ with $4$ in the expression.

$f (4) = 8 {(4)}^{2} - 64 \cdot 4 + 2$

Simplify the result.

Simplify each term.

Raise $4$ to the power of $2$ .

$f (4) = 8 \cdot 16 - 64 \cdot 4 + 2$

Multiply $8$ by $16$ .

$f (4) = 128 - 64 \cdot 4 + 2$

Multiply $- 64$ by $4$ .

$f (4) = 128 - 256 + 2$

Simplify by adding and subtracting.

Subtract $256$ from $128$ .

$f (4) = - 128 + 2$

Add $- 128$ and $2$ .

$f (4) = - 126$

The final answer is $- 126$ .

$- 126$

Step 4

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

$(4, - 126)$

Step 5

Find the Maximum/Minimum Value f(x)=8x^2-64x+2

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