Find the Maximum/Minimum Value f(x)=-0.02x^2+1.4x+5.6

$f (x) = - 0.02 x^{2} + 1.4 x + 5.6$

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 $b$ .

$x = - \frac{1.4}{2 (- 0.02)}$

Remove parentheses.

$x = - \frac{1.4}{2 (- 0.02)}$

Simplify $- \frac{1.4}{2 (- 0.02)}$ .

Multiply $2$ by $- 0.02$ .

$x = - \frac{1.4}{- 0.04}$

Divide $1.4$ by $- 0.04$ .

$x = - - 35$

Multiply $- 1$ by $- 35$ .

$x = 35$

Step 3

Evaluate $f (35)$ .

Replace the variable $x$ with $35$ in the expression.

$f (35) = - 0.02 {(35)}^{2} + 1.4 (35) + 5.6$

Simplify the result.

Simplify each term.

Raise $35$ to the power of $2$ .

$f (35) = - 0.02 \cdot 1225 + 1.4 (35) + 5.6$

Multiply $- 0.02$ by $1225$ .

$f (35) = - 24.5 + 1.4 (35) + 5.6$

Multiply $1.4$ by $35$ .

$f (35) = - 24.5 + 49 + 5.6$

Simplify by adding numbers.

Add $- 24.5$ and $49$ .

$f (35) = 24.5 + 5.6$

Add $24.5$ and $5.6$ .

$f (35) = 30.1$

The final answer is $30.1$ .

$30.1$

Step 4

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

$(35, 30.1)$

Step 5

Find the Maximum/Minimum Value f(x)=-0.02x^2+1.4x+5.6

Related Questions

Support LCMGCF.com by disabling your adblocker.