Find the Maximum/Minimum Value 3x^2+18x-4

$3 x^{2} + 18 x - 4$

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 $b$ .

$x = - \frac{18}{2 (3)}$

Remove parentheses.

$x = - \frac{18}{2 (3)}$

Simplify $- \frac{18}{2 (3)}$ .

Cancel the common factor of $18$ and $2$ .

Factor $2$ out of $18$ .

$x = - \frac{2 \cdot 9}{2 \cdot 3}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 9}{2 (3)}$

Cancel the common factor.

$x = - \frac{2 \cdot 9}{2 \cdot 3}$

Rewrite the expression.

$x = - \frac{9}{3}$

Cancel the common factor of $9$ and $3$ .

Factor $3$ out of $9$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3$ by $1$ .

$x = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x = - 3$

Step 3

Evaluate $f (- 3)$ .

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

$f (- 3) = 3 {(- 3)}^{2} + 18 (- 3) - 4$

Simplify the result.

Simplify each term.

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

$f (- 3) = 3 \cdot 9 + 18 (- 3) - 4$

Multiply $3$ by $9$ .

$f (- 3) = 27 + 18 (- 3) - 4$

Multiply $18$ by $- 3$ .

$f (- 3) = 27 - 54 - 4$

Simplify by subtracting numbers.

Subtract $54$ from $27$ .

$f (- 3) = - 27 - 4$

Subtract $4$ from $- 27$ .

$f (- 3) = - 31$

The final answer is $- 31$ .

$- 31$

Step 4

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

$(- 3, - 31)$

Step 5

Find the Maximum/Minimum Value 3x^2+18x-4

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