Find the Maximum/Minimum Value f(x)=4x^2-16x+6

$f (x) = 4 x^{2} - 16 x + 6$

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 $b$ .

$x = - \frac{- 16}{2 (4)}$

Remove parentheses.

$x = - \frac{- 16}{2 (4)}$

Simplify $- \frac{- 16}{2 (4)}$ .

Cancel the common factor of $- 16$ and $2$ .

Factor $2$ out of $- 16$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $4$ out of $- 8$ .

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

Cancel the common factors.

Factor $4$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 2$ by $1$ .

$x = - - 2$

Multiply $- 1$ by $- 2$ .

$x = 2$

Evaluate $f (2)$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = 4 {(2)}^{2} - 16 \cdot 2 + 6$

Simplify the result.

Simplify each term.

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

$f (2) = 4 \cdot 4 - 16 \cdot 2 + 6$

Multiply $4$ by $4$ .

$f (2) = 16 - 16 \cdot 2 + 6$

Multiply $- 16$ by $2$ .

$f (2) = 16 - 32 + 6$

Simplify by adding and subtracting.

Subtract $32$ from $16$ .

$f (2) = - 16 + 6$

Add $- 16$ and $6$ .

$f (2) = - 10$

The final answer is $- 10$ .

$- 10$

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

$(2, - 10)$

Find the Maximum/Minimum Value f(x)=4x^2-16x+6

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