Find the Maximum/Minimum Value y=3x^2+12x-4

$y = 3 x^{2} + 12 x - 4$

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 $b$ .

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

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 $6$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

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.

$f (- 2) = 3 {(- 2)}^{2} + 12 (- 2) - 4$

Simplify the result.

Simplify each term.

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

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

Multiply $3$ by $4$ .

$f (- 2) = 12 + 12 (- 2) - 4$

Multiply $12$ by $- 2$ .

$f (- 2) = 12 - 24 - 4$

Simplify by subtracting numbers.

Subtract $24$ from $12$ .

$f (- 2) = - 12 - 4$

Subtract $4$ from $- 12$ .

$f (- 2) = - 16$

The final answer is $- 16$ .

$- 16$

Step 4

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

$(- 2, - 16)$

Step 5

Find the Maximum/Minimum Value y=3x^2+12x-4

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