Find the Maximum/Minimum Value p(x)=-5x^2+260x-2975

$p (x) = - 5 x^{2} + 260 x - 2975$

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 $b$ .

$x = - \frac{260}{2 (- 5)}$

Remove parentheses.

$x = - \frac{260}{2 (- 5)}$

Simplify $- \frac{260}{2 (- 5)}$ .

Cancel the common factor of $260$ and $2$ .

Factor $2$ out of $260$ .

$x = - \frac{2 \cdot 130}{2 \cdot - 5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 5$ .

$x = - \frac{2 \cdot 130}{2 (- 5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 130}{2 \cdot - 5}$

Rewrite the expression.

$x = - \frac{130}{- 5}$

Cancel the common factor of $130$ and $- 5$ .

Factor $5$ out of $130$ .

$x = - \frac{5 (26)}{- 5}$

Move the negative one from the denominator of $\frac{26}{- 1}$ .

$x = - (- 1 \cdot 26)$

Multiply $- (- 1 \cdot 26)$ .

Multiply $- 1$ by $26$ .

$x = - - 26$

Multiply $- 1$ by $- 26$ .

$x = 26$

Step 3

Evaluate $f (26)$ .

Replace the variable $x$ with $26$ in the expression.

$p (26) = - 5 {(26)}^{2} + 260 (26) - 2975$

Simplify the result.

Simplify each term.

Raise $26$ to the power of $2$ .

$p (26) = - 5 \cdot 676 + 260 (26) - 2975$

Multiply $- 5$ by $676$ .

$p (26) = - 3380 + 260 (26) - 2975$

Multiply $260$ by $26$ .

$p (26) = - 3380 + 6760 - 2975$

Simplify by adding and subtracting.

Add $- 3380$ and $6760$ .

$p (26) = 3380 - 2975$

Subtract $2975$ from $3380$ .

$p (26) = 405$

The final answer is $405$ .

$405$

Step 4

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

$(26, 405)$

Step 5

Find the Maximum/Minimum Value p(x)=-5x^2+260x-2975

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