Precalculus Examples

$S (t) = t^{2} + 3 t - 6$ , $t = 0$ , $t = 2$

Step 1

Substitute using the average rate of change formula.

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Step 1.1

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 $t$ values of the two points.

$\frac{f (2) - f (0)}{(2) - (0)}$

Step 1.2

Substitute the equation $y = t^{2} + 3 t - 6$ for $f (2)$ and $f (0)$ , replacing $t$ in the function with the corresponding $t$ value.

$\frac{({(2)}^{2} + 3 (2) - 6) - ({(0)}^{2} + 3 (0) - 6)}{(2) - (0)}$

Step 2

Simplify the expression.

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Step 2.1

Simplify the numerator.

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Step 2.1.1

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

$\frac{4 + 3 (2) - 6 - (0^{2} + 3 (0) - 6)}{2 - (0)}$

Step 2.1.2

Multiply $3$ by $2$ .

$\frac{4 + 6 - 6 - (0^{2} + 3 (0) - 6)}{2 - (0)}$

Step 2.1.3

Simplify each term.

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Step 2.1.3.1

Raising $0$ to any positive power yields $0$ .

$\frac{4 + 6 - 6 - (0 + 3 (0) - 6)}{2 - (0)}$

Step 2.1.3.2

Multiply $3$ by $0$ .

$\frac{4 + 6 - 6 - (0 + 0 - 6)}{2 - (0)}$

Step 2.1.4

Add $0$ and $0$ .

$\frac{4 + 6 - 6 - (0 - 6)}{2 - (0)}$

Step 2.1.5

Subtract $6$ from $0$ .

$\frac{4 + 6 - 6 - - 6}{2 - (0)}$

Step 2.1.6

Multiply $- 1$ by $- 6$ .

$\frac{4 + 6 - 6 + 6}{2 - (0)}$

Step 2.1.7

Add $4$ and $6$ .

$\frac{10 - 6 + 6}{2 - (0)}$

Step 2.1.8

Subtract $6$ from $10$ .

$\frac{4 + 6}{2 - (0)}$

Step 2.1.9

Add $4$ and $6$ .

$\frac{10}{2 - (0)}$

Step 2.2

Simplify the denominator.

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Step 2.2.1

Multiply $- 1$ by $0$ .

$\frac{10}{2 + 0}$

Step 2.2.2

Add $2$ and $0$ .

$\frac{10}{2}$

Step 2.3

Divide $10$ by $2$ .

$5$