Finite Math Examples

$f (x) = - 0.16 x^{2} + 0.32 x + 1.16$

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 $x = - \frac{b}{2 a}$ .

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Step 2.1

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

$x = - \frac{0.32}{2 (- 0.16)}$

Step 2.2

Remove parentheses.

$x = - \frac{0.32}{2 (- 0.16)}$

Step 2.3

Simplify $- \frac{0.32}{2 (- 0.16)}$ .

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Step 2.3.1

Multiply $2$ by $- 0.16$ .

$x = - \frac{0.32}{- 0.32}$

Step 2.3.2

Divide $0.32$ by $- 0.32$ .

$x = - - 1$

Step 2.3.3

Multiply $- 1$ by $- 1$ .

$x = 1$

Step 3

Evaluate $f (1)$ .

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Step 3.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = - 0.16 {(1)}^{2} + 0.32 (1) + 1.16$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

One to any power is one.

$f (1) = - 0.16 \cdot 1 + 0.32 (1) + 1.16$

Step 3.2.1.2

Multiply $- 0.16$ by $1$ .

$f (1) = - 0.16 + 0.32 (1) + 1.16$

Step 3.2.1.3

Multiply $0.32$ by $1$ .

$f (1) = - 0.16 + 0.32 + 1.16$

Step 3.2.2

Simplify by adding numbers.

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Step 3.2.2.1

Add $- 0.16$ and $0.32$ .

$f (1) = 0.16 + 1.16$

Step 3.2.2.2

Add $0.16$ and $1.16$ .

$f (1) = 1.32$

Step 3.2.3

The final answer is $1.32$ .

$1.32$

Step 4

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

$(1, 1.32)$

Step 5