Find the Maximum/Minimum Value y=-2x^2+12x-19

$y = - 2 x^{2} + 12 x - 19$

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 $b$ .

$x = - \frac{12}{2 (- 2)}$

Remove parentheses.

$x = - \frac{12}{2 (- 2)}$

Simplify $- \frac{12}{2 (- 2)}$ .

Cancel the common factor of $12$ and $2$ .

Factor $2$ out of $12$ .

$x = - \frac{2 \cdot 6}{2 \cdot - 2}$

Cancel the common factors.

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

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

Cancel the common factor.

$x = - \frac{2 \cdot 6}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{6}{- 2}$

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

Factor $2$ out of $6$ .

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

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$x = - (- 1 \cdot 3)$

Multiply $- (- 1 \cdot 3)$ .

Multiply $- 1$ by $3$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = - 2 {(3)}^{2} + 12 (3) - 19$

Simplify the result.

Simplify each term.

Raise $3$ to the power of $2$ .

$f (3) = - 2 \cdot 9 + 12 (3) - 19$

Multiply $- 2$ by $9$ .

$f (3) = - 18 + 12 (3) - 19$

Multiply $12$ by $3$ .

$f (3) = - 18 + 36 - 19$

Simplify by adding and subtracting.

Add $- 18$ and $36$ .

$f (3) = 18 - 19$

Subtract $19$ from $18$ .

$f (3) = - 1$

The final answer is $- 1$ .

$- 1$

Step 4

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

$(3, - 1)$

Step 5

Find the Maximum/Minimum Value y=-2x^2+12x-19

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