Find the Maximum/Minimum Value f(x)=-0.1x^2+1.2x+5

$f (x) = - 0.1 x^{2} + 1.2 x + 5$

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 $b$ .

$x = - \frac{1.2}{2 (- 0.1)}$

Remove parentheses.

$x = - \frac{1.2}{2 (- 0.1)}$

Simplify $- \frac{1.2}{2 (- 0.1)}$ .

Multiply $2$ by $- 0.1$ .

$x = - \frac{1.2}{- 0.2}$

Divide $1.2$ by $- 0.2$ .

$x = - - 6$

Multiply $- 1$ by $- 6$ .

$x = 6$

Step 3

Evaluate $f (6)$ .

Replace the variable $x$ with $6$ in the expression.

$f (6) = - 0.1 {(6)}^{2} + 1.2 (6) + 5$

Simplify the result.

Simplify each term.

Raise $6$ to the power of $2$ .

$f (6) = - 0.1 \cdot 36 + 1.2 (6) + 5$

Multiply $- 0.1$ by $36$ .

$f (6) = - 3.6 + 1.2 (6) + 5$

Multiply $1.2$ by $6$ .

$f (6) = - 3.6 + 7.2 + 5$

Simplify by adding numbers.

Add $- 3.6$ and $7.2$ .

$f (6) = 3.6 + 5$

Add $3.6$ and $5$ .

$f (6) = 8.6$

The final answer is $8.6$ .

$8.6$

Step 4

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

$(6, 8.6)$

Step 5

Find the Maximum/Minimum Value f(x)=-0.1x^2+1.2x+5

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