Find the Maximum/Minimum Value f(x)=-0.2x^2+2.1x+6

$f (x) = - 0.2 x^{2} + 2.1 x + 6$

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 $b$ .

$x = - \frac{2.1}{2 (- 0.2)}$

Remove parentheses.

$x = - \frac{2.1}{2 (- 0.2)}$

Simplify $- \frac{2.1}{2 (- 0.2)}$ .

Multiply $2$ by $- 0.2$ .

$x = - \frac{2.1}{- 0.4}$

Divide $2.1$ by $- 0.4$ .

$x = - - 5.25$

Multiply $- 1$ by $- 5.25$ .

$x = 5.25$

Step 3

Evaluate $f (5.25)$ .

Replace the variable $x$ with $5.25$ in the expression.

$f (5.25) = - 0.2 {(5.25)}^{2} + 2.1 (5.25) + 6$

Simplify the result.

Simplify each term.

Raise $5.25$ to the power of $2$ .

$f (5.25) = - 0.2 \cdot 27.5625 + 2.1 (5.25) + 6$

Multiply $- 0.2$ by $27.5625$ .

$f (5.25) = - 5.5125 + 2.1 (5.25) + 6$

Multiply $2.1$ by $5.25$ .

$f (5.25) = - 5.5125 + 11.025 + 6$

Simplify by adding numbers.

Add $- 5.5125$ and $11.025$ .

$f (5.25) = 5.5125 + 6$

Add $5.5125$ and $6$ .

$f (5.25) = 11.5125$

The final answer is $11.5125$ .

$11.5125$

Step 4

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

$(5.25, 11.5125)$

Step 5

Find the Maximum/Minimum Value f(x)=-0.2x^2+2.1x+6

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