Find the Maximum/Minimum Value f(x)=3x^2+6x-72

$f (x) = 3 x^{2} + 6 x - 72$

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 $b$ .

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

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 $3$ .

Cancel the common factor.

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

Rewrite the expression.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

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

$f (- 1) = 3 {(- 1)}^{2} + 6 (- 1) - 72$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = 3 \cdot 1 + 6 (- 1) - 72$

Multiply $3$ by $1$ .

$f (- 1) = 3 + 6 (- 1) - 72$

Multiply $6$ by $- 1$ .

$f (- 1) = 3 - 6 - 72$

Simplify by subtracting numbers.

Subtract $6$ from $3$ .

$f (- 1) = - 3 - 72$

Subtract $72$ from $- 3$ .

$f (- 1) = - 75$

The final answer is $- 75$ .

$- 75$

Step 4

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

$(- 1, - 75)$

Step 5

Find the Maximum/Minimum Value f(x)=3x^2+6x-72

Related Questions

Support LCMGCF.com by disabling your adblocker.