Precalculus Examples

$h (t) = {(t + 3)}^{2} + 5$ $- 5 \leq t \leq - 1$

Step 1

Write $h (t) = {(t + 3)}^{2} + 5$ as an equation.

$y = {(t + 3)}^{2} + 5$

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 $t$ values of the two points.

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

Step 2.2

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

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Add $- 1$ and $3$ .

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

Step 3.1.2

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

$\frac{4 + 5 - ({(- 5 + 3)}^{2} + 5)}{- 1 - (- 5)}$

Step 3.1.3

Simplify each term.

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Step 3.1.3.1

Add $- 5$ and $3$ .

$\frac{4 + 5 - ({(- 2)}^{2} + 5)}{- 1 - (- 5)}$

Step 3.1.3.2

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

$\frac{4 + 5 - (4 + 5)}{- 1 - (- 5)}$

Step 3.1.4

Add $4$ and $5$ .

$\frac{4 + 5 - 1 \cdot 9}{- 1 - (- 5)}$

Step 3.1.5

Multiply $- 1$ by $9$ .

$\frac{4 + 5 - 9}{- 1 - (- 5)}$

Step 3.1.6

Add $4$ and $5$ .

$\frac{9 - 9}{- 1 - (- 5)}$

Step 3.1.7

Subtract $9$ from $9$ .

$\frac{0}{- 1 - (- 5)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $- 5$ .

$\frac{0}{- 1 + 5}$

Step 3.2.2

Add $- 1$ and $5$ .

$\frac{0}{4}$

Step 3.3

Divide $0$ by $4$ .

$0$