Finite Math Examples

$f (x) = 9.6 x^{2} - 3 x - 5$ , $(1, 5)$

Step 1

Write $f (x) = 9.6 x^{2} - 3 x - 5$ as an equation.

$y = 9.6 x^{2} - 3 x - 5$

Step 2

Substitute using the average rate of change formula.

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Step 2.1

The average rate of change of a function can be found by calculating the change in $y$ values of the two points divided by the change in $x$ values of the two points.

$\frac{f (5) - f (1)}{(5) - (1)}$

Step 2.2

Substitute the equation $y = 9.6 x^{2} - 3 x - 5$ for $f (5)$ and $f (1)$ , replacing $x$ in the function with the corresponding $x$ value.

$\frac{(9.6 {(5)}^{2} - 3 \cdot 5 - 5) - (9.6 {(1)}^{2} - 3 \cdot 1 - 5)}{(5) - (1)}$

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $5$ to the power of $2$ .

$\frac{9.6 \cdot 25 - 3 \cdot 5 - 5 - (9.6 \cdot 1^{2} - 3 \cdot 1 - 5)}{5 - (1)}$

Step 3.1.2

Multiply $9.6$ by $25$ .

$\frac{240 - 3 \cdot 5 - 5 - (9.6 \cdot 1^{2} - 3 \cdot 1 - 5)}{5 - (1)}$

Step 3.1.3

Multiply $- 3$ by $5$ .

$\frac{240 - 15 - 5 - (9.6 \cdot 1^{2} - 3 \cdot 1 - 5)}{5 - (1)}$

Step 3.1.4

Simplify each term.

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Step 3.1.4.1

One to any power is one.

$\frac{240 - 15 - 5 - (9.6 \cdot 1 - 3 \cdot 1 - 5)}{5 - (1)}$

Step 3.1.4.2

Multiply $9.6$ by $1$ .

$\frac{240 - 15 - 5 - (9.6 - 3 \cdot 1 - 5)}{5 - (1)}$

Step 3.1.4.3

Multiply $- 3$ by $1$ .

$\frac{240 - 15 - 5 - (9.6 - 3 - 5)}{5 - (1)}$

Step 3.1.5

Subtract $3$ from $9.6$ .

$\frac{240 - 15 - 5 - (6.6 - 5)}{5 - (1)}$

Step 3.1.6

Subtract $5$ from $6.6$ .

$\frac{240 - 15 - 5 - 1 \cdot 1.6}{5 - (1)}$

Step 3.1.7

Multiply $- 1$ by $1.6$ .

$\frac{240 - 15 - 5 - 1.6}{5 - (1)}$

Step 3.1.8

Subtract $15$ from $240$ .

$\frac{225 - 5 - 1.6}{5 - (1)}$

Step 3.1.9

Subtract $5$ from $225$ .

$\frac{220 - 1.6}{5 - (1)}$

Step 3.1.10

Subtract $1.6$ from $220$ .

$\frac{218.4}{5 - (1)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $1$ .

$\frac{218.4}{5 - 1}$

Step 3.2.2

Subtract $1$ from $5$ .

$\frac{218.4}{4}$

Step 3.3

Divide $218.4$ by $4$ .

$54.6$