Find the Maximum/Minimum Value y=-6x^2+36x+18

$y = - 6 x^{2} + 36 x + 18$

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 $b$ .

$x = - \frac{36}{2 (- 6)}$

Remove parentheses.

$x = - \frac{36}{2 (- 6)}$

Simplify $- \frac{36}{2 (- 6)}$ .

Cancel the common factor of $36$ and $2$ .

Factor $2$ out of $36$ .

$x = - \frac{2 \cdot 18}{2 \cdot - 6}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 6$ .

$x = - \frac{2 \cdot 18}{2 (- 6)}$

Cancel the common factor.

$x = - \frac{2 \cdot 18}{2 \cdot - 6}$

Rewrite the expression.

$x = - \frac{18}{- 6}$

Cancel the common factor of $18$ and $- 6$ .

Factor $6$ out of $18$ .

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

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

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

Multiply $- (- 1 \cdot 3)$ .

Multiply $- 1$ by $3$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

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

$f (3) = - 6 {(3)}^{2} + 36 (3) + 18$

Simplify the result.

Simplify each term.

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

$f (3) = - 6 \cdot 9 + 36 (3) + 18$

Multiply $- 6$ by $9$ .

$f (3) = - 54 + 36 (3) + 18$

Multiply $36$ by $3$ .

$f (3) = - 54 + 108 + 18$

Simplify by adding numbers.

Add $- 54$ and $108$ .

$f (3) = 54 + 18$

Add $54$ and $18$ .

$f (3) = 72$

The final answer is $72$ .

$72$

Step 4

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

$(3, 72)$

Step 5

Find the Maximum/Minimum Value y=-6x^2+36x+18

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