Find the Maximum/Minimum Value -3x^2+18x-31

$- 3 x^{2} + 18 x - 31$

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 $x = - \frac{b}{2 a}$

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 $9$ .

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

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

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

Multiply.

Multiply $- 1$ by $3$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Evaluate $f (3)$ .

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

$f (3) = - 3 {(3)}^{2} + 18 (3) - 31$

Simplify the result.

Simplify each term.

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

$f (3) = - 3 \cdot 9 + 18 (3) - 31$

Multiply $- 3$ by $9$ .

$f (3) = - 27 + 18 (3) - 31$

Multiply $18$ by $3$ .

$f (3) = - 27 + 54 - 31$

Simplify by adding and subtracting.

Add $- 27$ and $54$ .

$f (3) = 27 - 31$

Subtract $31$ from $27$ .

$f (3) = - 4$

The final answer is $- 4$ .

$- 4$

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

$(3, - 4)$

Find the Maximum/Minimum Value -3x^2+18x-31

Related Questions

Support LCMGCF.com by disabling your adblocker.