Find the Maximum/Minimum Value p(x)=-2x^2+40x-72

$p (x) = - 2 x^{2} + 40 x - 72$

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 $b$ .

$x = - \frac{40}{2 (- 2)}$

Remove parentheses.

$x = - \frac{40}{2 (- 2)}$

Simplify $- \frac{40}{2 (- 2)}$ .

Cancel the common factor of $40$ and $2$ .

Factor $2$ out of $40$ .

$x = - \frac{2 \cdot 20}{2 \cdot - 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 20}{2 (- 2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 20}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{20}{- 2}$

Cancel the common factor of $20$ and $- 2$ .

Factor $2$ out of $20$ .

$x = - \frac{2 (10)}{- 2}$

Move the negative one from the denominator of $\frac{10}{- 1}$ .

$x = - (- 1 \cdot 10)$

Multiply $- (- 1 \cdot 10)$ .

Multiply $- 1$ by $10$ .

$x = - - 10$

Multiply $- 1$ by $- 10$ .

$x = 10$

Step 3

Evaluate $f (10)$ .

Replace the variable $x$ with $10$ in the expression.

$p (10) = - 2 {(10)}^{2} + 40 (10) - 72$

Simplify the result.

Simplify each term.

Raise $10$ to the power of $2$ .

$p (10) = - 2 \cdot 100 + 40 (10) - 72$

Multiply $- 2$ by $100$ .

$p (10) = - 200 + 40 (10) - 72$

Multiply $40$ by $10$ .

$p (10) = - 200 + 400 - 72$

Simplify by adding and subtracting.

Add $- 200$ and $400$ .

$p (10) = 200 - 72$

Subtract $72$ from $200$ .

$p (10) = 128$

The final answer is $128$ .

$128$

Step 4

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

$(10, 128)$

Step 5

Find the Maximum/Minimum Value p(x)=-2x^2+40x-72

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