Find the Maximum/Minimum Value f(x)=6-x^2

$f (x) = 6 - x^{2}$

Step 1

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ 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 $0$ .

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

Move the negative one from the denominator of $\frac{0}{- 1}$ .

$x = - (- 1 \cdot 0)$

Rewrite $- 1 \cdot 0$ as $- 0$ .

$x = - - 0$

Multiply $- - 0$ .

Multiply $- 1$ by $0$ .

$x = - 0$

Multiply $- 1$ by $0$ .

$x = 0$

Step 3

Evaluate $f (0)$ .

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

$f (0) = 6 - {(0)}^{2}$

Simplify the result.

Simplify each term.

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

$f (0) = 6 - 0$

Multiply $- 1$ by $0$ .

$f (0) = 6 + 0$

Add $6$ and $0$ .

$f (0) = 6$

The final answer is $6$ .

$6$

Step 4

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

$(0, 6)$

Step 5

Find the Maximum/Minimum Value f(x)=6-x^2

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