Precalculus Examples

$h (t) = {(t + 3)}^{2} + 5$ $- 4 \leq t \leq - 3$

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.

$\frac{f (- 3) - f (- 4)}{(- 3) - (- 4)}$

Step 2.2

Substitute the equation $y = {(t + 3)}^{2} + 5$ for $f (- 3)$ and $f (- 4)$ , replacing $t$ in the function with the corresponding $t$ value.

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Add $- 3$ and $3$ .

$\frac{0^{2} + 5 - ({(- 4 + 3)}^{2} + 5)}{- 3 - (- 4)}$

Step 3.1.2

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

$\frac{0 + 5 - ({(- 4 + 3)}^{2} + 5)}{- 3 - (- 4)}$

Step 3.1.3

Simplify each term.

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Step 3.1.3.1

Add $- 4$ and $3$ .

$\frac{0 + 5 - ({(- 1)}^{2} + 5)}{- 3 - (- 4)}$

Step 3.1.3.2

Raise $- 1$ to the power of $2$ .

$\frac{0 + 5 - (1 + 5)}{- 3 - (- 4)}$

Step 3.1.4

Add $1$ and $5$ .

$\frac{0 + 5 - 1 \cdot 6}{- 3 - (- 4)}$

Step 3.1.5

Multiply $- 1$ by $6$ .

$\frac{0 + 5 - 6}{- 3 - (- 4)}$

Step 3.1.6

Add $0$ and $5$ .

$\frac{5 - 6}{- 3 - (- 4)}$

Step 3.1.7

Subtract $6$ from $5$ .

$\frac{- 1}{- 3 - (- 4)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $- 4$ .

$\frac{- 1}{- 3 + 4}$

Step 3.2.2

Add $- 3$ and $4$ .

$\frac{- 1}{1}$

Step 3.3

Divide $- 1$ by $1$ .

$- 1$