Find the Maximum/Minimum Value g(x)=x^2-6x+10

$g (x) = x^{2} - 6 x + 10$

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 $b$ .

$x = - \frac{- 6}{2 (1)}$

Remove parentheses.

$x = - \frac{- 6}{2 (1)}$

Simplify $- \frac{- 6}{2 (1)}$ .

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

$x = - \frac{2 \cdot - 3}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$x = - \frac{2 \cdot - 3}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 3}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{- 3}{1}$

Divide $- 3$ by $1$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

Replace the variable $x$ with $3$ in the expression.

$g (3) = {(3)}^{2} - 6 \cdot 3 + 10$

Simplify the result.

Simplify each term.

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

$g (3) = 9 - 6 \cdot 3 + 10$

Multiply $- 6$ by $3$ .

$g (3) = 9 - 18 + 10$

Simplify by adding and subtracting.

Subtract $18$ from $9$ .

$g (3) = - 9 + 10$

Add $- 9$ and $10$ .

$g (3) = 1$

The final answer is $1$ .

$1$

Step 4

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

$(3, 1)$

Step 5

Find the Maximum/Minimum Value g(x)=x^2-6x+10

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