Find the Maximum/Minimum Value f(x)=-16x^2+160x

$f (x) = - 16 x^{2} + 160 x$

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 $b$ .

$x = - \frac{160}{2 (- 16)}$

Remove parentheses.

$x = - \frac{160}{2 (- 16)}$

Simplify $- \frac{160}{2 (- 16)}$ .

Cancel the common factor of $160$ and $2$ .

Factor $2$ out of $160$ .

$x = - \frac{2 \cdot 80}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$x = - \frac{2 \cdot 80}{2 (- 16)}$

Cancel the common factor.

$x = - \frac{2 \cdot 80}{2 \cdot - 16}$

Rewrite the expression.

$x = - \frac{80}{- 16}$

Cancel the common factor of $80$ and $- 16$ .

Factor $16$ out of $80$ .

$x = - \frac{16 (5)}{- 16}$

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$x = - (- 1 \cdot 5)$

Multiply $- (- 1 \cdot 5)$ .

Multiply $- 1$ by $5$ .

$x = - - 5$

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 3

Evaluate $f (5)$ .

Replace the variable $x$ with $5$ in the expression.

$f (5) = - 16 {(5)}^{2} + 160 (5)$

Simplify the result.

Simplify each term.

Raise $5$ to the power of $2$ .

$f (5) = - 16 \cdot 25 + 160 (5)$

Multiply $- 16$ by $25$ .

$f (5) = - 400 + 160 (5)$

Multiply $160$ by $5$ .

$f (5) = - 400 + 800$

Add $- 400$ and $800$ .

$f (5) = 400$

The final answer is $400$ .

$400$

Step 4

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

$(5, 400)$

Step 5

Find the Maximum/Minimum Value f(x)=-16x^2+160x

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