Find the Maximum/Minimum Value f(x)=2x^2-8x+9

$f (x) = 2 x^{2} - 8 x + 9$

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 $b$ .

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

Remove parentheses.

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

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

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

Factor $2$ out of $- 8$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

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.

$f (2) = 2 {(2)}^{2} - 8 \cdot 2 + 9$

Simplify the result.

Simplify each term.

Multiply $2$ by ${(2)}^{2}$ by adding the exponents.

Multiply $2$ by ${(2)}^{2}$ .

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

$f (2) = 2 {(2)}^{2} - 8 \cdot 2 + 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (2) = 2^{1 + 2} - 8 \cdot 2 + 9$

Add $1$ and $2$ .

$f (2) = 2^{3} - 8 \cdot 2 + 9$

Raise $2$ to the power of $3$ .

$f (2) = 8 - 8 \cdot 2 + 9$

Multiply $- 8$ by $2$ .

$f (2) = 8 - 16 + 9$

Simplify by adding and subtracting.

Subtract $16$ from $8$ .

$f (2) = - 8 + 9$

Add $- 8$ and $9$ .

$f (2) = 1$

The final answer is $1$ .

$1$

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

$(2, 1)$

Find the Maximum/Minimum Value f(x)=2x^2-8x+9

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