Find the Maximum/Minimum Value x^2-9

$x^{2} - 9$

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 $b$ .

$x = - \frac{0}{2 (1)}$

Remove parentheses.

$x = - \frac{0}{2 (1)}$

Simplify $- \frac{0}{2 (1)}$ .

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$x = - \frac{2 (0)}{2 (1)}$

Cancel the common factors.

Cancel the common factor.

$x = - \frac{2 \cdot 0}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{0}{1}$

Divide $0$ by $1$ .

$x = - 0$

Multiply $- 1$ by $0$ .

$x = 0$

Step 3

Evaluate $f (0)$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{2} - 9$

Simplify the result.

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 - 9$

Subtract $9$ from $0$ .

$f (0) = - 9$

The final answer is $- 9$ .

$- 9$

Step 4

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

$(0, - 9)$

Step 5

Find the Maximum/Minimum Value x^2-9

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