Find the Maximum/Minimum Value f(x)=4x^2-8x+6

$f (x) = 4 x^{2} - 8 x + 6$

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 $- 4$ and $4$ .

Factor $4$ out of $- 4$ .

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

Cancel the common factors.

Factor $4$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$x = - - 1$

Multiply $- 1$ by $- 1$ .

$x = 1$

Step 3

Evaluate $f (1)$ .

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

$f (1) = 4 {(1)}^{2} - 8 \cdot 1 + 6$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 4 \cdot 1 - 8 \cdot 1 + 6$

Multiply $4$ by $1$ .

$f (1) = 4 - 8 \cdot 1 + 6$

Multiply $- 8$ by $1$ .

$f (1) = 4 - 8 + 6$

Simplify by adding and subtracting.

Subtract $8$ from $4$ .

$f (1) = - 4 + 6$

Add $- 4$ and $6$ .

$f (1) = 2$

The final answer is $2$ .

$2$

Step 4

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

$(1, 2)$

Step 5

Find the Maximum/Minimum Value f(x)=4x^2-8x+6

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