Find the Maximum/Minimum Value y=-0.275x^2+6.05x+260

$y = - 0.275 x^{2} + 6.05 x + 260$

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 $b$ .

$x = - \frac{6.05}{2 (- 0.275)}$

Remove parentheses.

$x = - \frac{6.05}{2 (- 0.275)}$

Simplify $- \frac{6.05}{2 (- 0.275)}$ .

Multiply $2$ by $- 0.275$ .

$x = - \frac{6.05}{- 0.55}$

Divide $6.05$ by $- 0.55$ .

$x = - - 11$

Multiply $- 1$ by $- 11$ .

$x = 11$

Step 3

Evaluate $f (11)$ .

Replace the variable $x$ with $11$ in the expression.

$f (11) = - 0.275 {(11)}^{2} + 6.05 (11) + 260$

Simplify the result.

Simplify each term.

Raise $11$ to the power of $2$ .

$f (11) = - 0.275 \cdot 121 + 6.05 (11) + 260$

Multiply $- 0.275$ by $121$ .

$f (11) = - 33.275 + 6.05 (11) + 260$

Multiply $6.05$ by $11$ .

$f (11) = - 33.275 + 66.55 + 260$

Simplify by adding numbers.

Add $- 33.275$ and $66.55$ .

$f (11) = 33.275 + 260$

Add $33.275$ and $260$ .

$f (11) = 293.275$

The final answer is $293.275$ .

$293.275$

Step 4

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

$(11, 293.275)$

Step 5

Find the Maximum/Minimum Value y=-0.275x^2+6.05x+260

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