Precalculus Examples

$h (t) = - 16 t^{2} + 64 t$ on $[2, 4]$

Step 1

Write $h (t) = - 16 t^{2} + 64 t$ as an equation.

$y = - 16 t^{2} + 64 t$

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.

$\frac{f (4) - f (2)}{(4) - (2)}$

Step 2.2

Substitute the equation $y = - 16 t^{2} + 64 t$ for $f (4)$ and $f (2)$ , replacing $t$ in the function with the corresponding $t$ value.

$\frac{(- 16 {(4)}^{2} + 64 (4)) - (- 16 {(2)}^{2} + 64 (2))}{(4) - (2)}$

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $4$ to the power of $2$ .

$\frac{- 16 \cdot 16 + 64 (4) - (- 16 \cdot 2^{2} + 64 (2))}{4 - (2)}$

Step 3.1.2

Multiply $- 16$ by $16$ .

$\frac{- 256 + 64 (4) - (- 16 \cdot 2^{2} + 64 (2))}{4 - (2)}$

Step 3.1.3

Multiply $64$ by $4$ .

$\frac{- 256 + 256 - (- 16 \cdot 2^{2} + 64 (2))}{4 - (2)}$

Step 3.1.4

Simplify each term.

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Step 3.1.4.1

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

$\frac{- 256 + 256 - (- 16 \cdot 4 + 64 (2))}{4 - (2)}$

Step 3.1.4.2

Multiply $- 16$ by $4$ .

$\frac{- 256 + 256 - (- 64 + 64 (2))}{4 - (2)}$

Step 3.1.4.3

Multiply $64$ by $2$ .

$\frac{- 256 + 256 - (- 64 + 128)}{4 - (2)}$

Step 3.1.5

Add $- 64$ and $128$ .

$\frac{- 256 + 256 - 1 \cdot 64}{4 - (2)}$

Step 3.1.6

Multiply $- 1$ by $64$ .

$\frac{- 256 + 256 - 64}{4 - (2)}$

Step 3.1.7

Add $- 256$ and $256$ .

$\frac{0 - 64}{4 - (2)}$

Step 3.1.8

Subtract $64$ from $0$ .

$\frac{- 64}{4 - (2)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $2$ .

$\frac{- 64}{4 - 2}$

Step 3.2.2

Subtract $2$ from $4$ .

$\frac{- 64}{2}$

Step 3.3

Divide $- 64$ by $2$ .

$- 32$