Find the Maximum/Minimum Value y=-1/4x^2+5x

$y = - \frac{1}{4} x^{2} + 5 x$

Step 1

Combine $x^{2}$ and $\frac{1}{4}$ .

$f (x) = - \frac{x^{2}}{4} + 5 x$

Step 2

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 3

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{5}{2 (- 0.25)}$

Remove parentheses.

$x = - \frac{5}{2 (- 0.25)}$

Simplify $- \frac{5}{2 (- 0.25)}$ .

Multiply $2$ by $- 0.25$ .

$x = - \frac{5}{- 0.5}$

Divide $5$ by $- 0.5$ .

$x = - - 10$

Multiply $- 1$ by $- 10$ .

$x = 10$

Step 4

Evaluate $f (10)$ .

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

$f (10) = - \frac{{(10)}^{2}}{4} + 5 (10)$

Simplify the result.

Simplify each term.

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

$f (10) = - \frac{100}{4} + 5 (10)$

Divide $100$ by $4$ .

$f (10) = - 1 \cdot 25 + 5 (10)$

Multiply $- 1$ by $25$ .

$f (10) = - 25 + 5 (10)$

Multiply $5$ by $10$ .

$f (10) = - 25 + 50$

Add $- 25$ and $50$ .

$f (10) = 25$

The final answer is $25$ .

$25$

Step 5

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

$(10, 25)$

Step 6

Find the Maximum/Minimum Value y=-1/4x^2+5x

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