Find the Maximum/Minimum Value y=x^2+10x+28

$y = x^{2} + 10 x + 28$

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 $1$ .

$x = - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$x = - 5$

Step 3

Evaluate $f (- 5)$ .

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

$f (- 5) = {(- 5)}^{2} + 10 (- 5) + 28$

Simplify the result.

Simplify each term.

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

$f (- 5) = 25 + 10 (- 5) + 28$

Multiply $10$ by $- 5$ .

$f (- 5) = 25 - 50 + 28$

Simplify by adding and subtracting.

Subtract $50$ from $25$ .

$f (- 5) = - 25 + 28$

Add $- 25$ and $28$ .

$f (- 5) = 3$

The final answer is $3$ .

$3$

Step 4

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

$(- 5, 3)$

Step 5

Find the Maximum/Minimum Value y=x^2+10x+28

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