Find the Maximum/Minimum Value y=-3x^2-12x+5

$y = - 3 x^{2} - 12 x + 5$

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 $b$ .

$x = - \frac{- 12}{2 (- 3)}$

Remove parentheses.

$x = - \frac{- 12}{2 (- 3)}$

Simplify $- \frac{- 12}{2 (- 3)}$ .

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

Factor $2$ out of $- 12$ .

$x = - \frac{2 \cdot - 6}{2 \cdot - 3}$

Cancel the common factors.

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

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

Cancel the common factor.

$x = - \frac{2 \cdot - 6}{2 \cdot - 3}$

Rewrite the expression.

$x = - \frac{- 6}{- 3}$

Cancel the common factor of $- 6$ and $- 3$ .

Factor $- 3$ out of $- 6$ .

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

Cancel the common factors.

Factor $- 3$ out of $- 3$ .

$x = - \frac{- 3 \cdot 2}{- 3 \cdot 1}$

Cancel the common factor.

$x = - \frac{- 3 \cdot 2}{- 3 \cdot 1}$

Rewrite the expression.

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

Divide $2$ by $1$ .

$x = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$x = - 2$

Step 3

Evaluate $f (- 2)$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = - 3 {(- 2)}^{2} - 12 \cdot - 2 + 5$

Simplify the result.

Simplify each term.

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

$f (- 2) = - 3 \cdot 4 - 12 \cdot - 2 + 5$

Multiply $- 3$ by $4$ .

$f (- 2) = - 12 - 12 \cdot - 2 + 5$

Multiply $- 12$ by $- 2$ .

$f (- 2) = - 12 + 24 + 5$

Simplify by adding numbers.

Add $- 12$ and $24$ .

$f (- 2) = 12 + 5$

Add $12$ and $5$ .

$f (- 2) = 17$

The final answer is $17$ .

$17$

Step 4

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

$(- 2, 17)$

Step 5

Find the Maximum/Minimum Value y=-3x^2-12x+5

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