Find the Maximum/Minimum Value y=x^2-10x+31

$y = x^{2} - 10 x + 31$

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 $b$ .

$x = - \frac{- 10}{2 (1)}$

Remove parentheses.

$x = - \frac{- 10}{2 (1)}$

Simplify $- \frac{- 10}{2 (1)}$ .

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10$ .

$x = - \frac{2 \cdot - 5}{2 \cdot 1}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 5}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 5}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{- 5}{1}$

Divide $- 5$ by $1$ .

$x = - - 5$

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 3

Evaluate $f (5)$ .

Replace the variable $x$ with $5$ in the expression.

$f (5) = {(5)}^{2} - 10 \cdot 5 + 31$

Simplify the result.

Simplify each term.

Raise $5$ to the power of $2$ .

$f (5) = 25 - 10 \cdot 5 + 31$

Multiply $- 10$ by $5$ .

$f (5) = 25 - 50 + 31$

Simplify by adding and subtracting.

Subtract $50$ from $25$ .

$f (5) = - 25 + 31$

Add $- 25$ and $31$ .

$f (5) = 6$

The final answer is $6$ .

$6$

Step 4

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

$(5, 6)$

Step 5

Find the Maximum/Minimum Value y=x^2-10x+31

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