Find the Maximum/Minimum Value -3x^2+6x+9

$- 3 x^{2} + 6 x + 9$

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 $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{6}{2 (- 3)}$

Remove parentheses.

$x = - \frac{6}{2 (- 3)}$

Simplify $- \frac{6}{2 (- 3)}$ .

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

Factor $2$ out of $6$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot - 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ and $- 3$ .

Factor $3$ out of $3$ .

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

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

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

Multiply.

Multiply $- 1$ by $1$ .

$x = - - 1$

Multiply $- 1$ by $- 1$ .

$x = 1$

Evaluate $f (1)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = - 3 {(1)}^{2} + 6 (1) + 9$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = - 3 \cdot 1 + 6 (1) + 9$

Multiply $- 3$ by $1$ .

$f (1) = - 3 + 6 (1) + 9$

Multiply $6$ by $1$ .

$f (1) = - 3 + 6 + 9$

Simplify by adding numbers.

Add $- 3$ and $6$ .

$f (1) = 3 + 9$

Add $3$ and $9$ .

$f (1) = 12$

The final answer is $12$ .

$12$

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

$(1, 12)$

Find the Maximum/Minimum Value -3x^2+6x+9

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