Average Rate of Change and Slope

4.8.2 Average Rate of Change and Slope

4.8.2 • Average Rate of Change and Slope

Activity

The table and graph show a more complete picture of the temperature changes on the same winter day. The function $T$ gives the temperature, in degrees Fahrenheit, $h$ hours since noon.

A scatter plot showing temperature in °F versus hours since noon. Points start at 20°F at 0 hours, peak at 25°F at 4 hours, then drop to 8°F at 10 hours. There is a data point for every hour since noon.

$h$	0	1	2	3	4	5	6	7	8	9	10	11	12
$T (h)$	18	19	20	20	25	23	17	15	11	11	8	6	7

Values of the function T(h) for integer inputs of h from 0 to 12.

In questions 1 – 3, find the average rate of change for the following intervals. Be prepared to show your reasoning.

1. between noon and 1 p.m.

Compare your answer:

1°F per hour. The temperature rose by 1°F in 1 hour.

2. between noon and 4 p.m.

Compare your answer:

1.75°F per hour. The temperature increased by 7°F in 4 hours. The average increase is $\frac{7}{4}$ or 1.75.

3. between noon and midnight

Compare your answer:

–0.92°F an hour. I found the slope of the line connecting the two points, and it is $- \frac{11}{12}$ .

In questions 4 – 6, use the slope formula to find the slope of the line that connects the following points.

4. $(0, 18)$ and $(1, 19)$

Compare your answer:

$\frac{19 - 18}{1 - 0} = 1$

5. $(0, 18)$ and $(4, 25)$

Compare your answer:

$\frac{25 - 18}{4 - 0} = \frac{7}{4}$

6. $(0, 18)$ and $(12, 7)$

Compare your answer:

$\frac{7 - 18}{12 - 0} = \frac{- 11}{12}$

7. Compare your answers to questions 1 – 3 with your answers to questions 4 – 6. Explain what this tells you about the average rate of change and the slope of a line.

Compare your answer:

The average rate of change between two points is the same as the slope of the line connecting the 2 points.

For questions 8 and 9, think back to Mai and Tyler’s disagreement about temperature to solve these problems.

8. Use the average rate of change to show which time period—4 p.m. to 6 p.m. or 6 p.m. to 10 p.m.—experienced a faster temperature drop.

Compare your answer:

The temperature dropped faster between 4 p.m. and 6 p.m. For example: In the first interval, the temperature fell 8°F in 2 hours, which is an average drop of 4°F an hour. In the second, it fell 9°F in 4 hours, which is an average drop of 2.25°F per hour.

9. Use the slope formula to show which line has the greater slope—the line connecting the points that represent the temperature at 4 p.m. and 6 p.m. or the line connecting the points that represent the temperature at 6 p.m. and 10 p.m.

Compare your answer:

The slope of the line representing the temperature between 4 p.m. and 6 p.m. has a greater slope. The slope of the line connecting the points representing the temperature between 4 p.m. and 6 p.m. is $\frac{17 - 25}{6 - 4} = - 4$ . The slope of the line connecting the points representing the temperature between 6 p.m. and 10 p.m. is $\frac{8 - 17}{10 - 6} = \frac{9}{4} = - 2.25$ .

10. Compare your answers to question 8 and question 9.

Compare your answer:

Finding the average rate of change and using the slope formula give the same results. The average rate of change and the slope of the line between hours 4 and 6 is greater than the average rate of change and the slope of the line between hours 6 and 10.

Are you ready for more?

Extending Your Thinking

Refer to graph and table of the previous function $T$ to answer the questions:

1. Over what interval did the temperature decrease the most rapidly?

Compare your answer:

between 5 p.m. and 6 p.m.

2. Over what interval did the temperature increase the most rapidly?

Compare your answer:

between 3 p.m. and 4 p.m.

Self Check

The temperature $t$ , in degrees Fahrenheit, of a cup of coffee that was poured and then forgotten for $m$ minutes is represented in the table below

Time in minutes	0	15	25	40
Temperature	185°F	155°F	140°F	120°F

What was the average rate of change from when the cup was poured to 40 minutes later?

-1.625 degrees per minute
-1.333 degrees per minute
-0.615 degrees per minute
-2 degrees per minute

Additional Resources

Finding Rates of Change

Jay was racing bikes with a friend on the way to the park. A table and a graph of the times are shown below.

Time (minutes)	Distance (miles)
0	0
5	0.84
10	1.86
15	3.00
20	4.27
25	5.67

Jay

A scatterplot that shows the relationship between time in minutes (x-axis) and distance in miles (y-axis) plotted points include (0, 0), (5, 0.84), (10, 1..86), (15, 3), (20, 42.7), and (25, 5.67).

What is the average rate of change of the first 10 minutes of the race?

To find the average rate of change of the first 10 minutes of the race, you would look at the first data value on the table and the third data value on the table, since it corresponds to 10 minutes.

Remember, the average rate of change formula is $\frac{f (b) - f (a)}{b - a}$ , where $(b, f (b))$ typically represents your second data point and $(a, f (a))$ represents your first data point.

For this situation, it would be $(0, 0)$ or $a = 0$ and $f (a) = 0$ and $(10, 1.86)$ or $b = 10$ and $f (b) = 1.86$ .

Substituting the points into the average rate of change formula results in the following:

$\frac{f (b) - f (a)}{b - a}$ , which is $\frac{1.86}{10}$ or 0.186 miles per minute.

Notice that the answer is positive and the points on the graph are higher each time.

Try it

Finding Rates of Change

Jay was racing bikes with a friend on the way to the park. A table and a graph of the times are shown below.

Time (minutes)	Distance (miles)
0	0
5	0.84
10	1.86
15	3.00
20	4.27
25	5.67

Jay

What was Jay’s average rate of change from the beginning to 25 minutes later?

Compare your answer:

Here is how to find the average rate of change of the first 25 minutes of the race:

Remember, the average rate of change formula is $\frac{f (b) - f (a)}{b - a}$ , where $(b, f (b))$ typically represents your second data point and $(a, f (a))$ represents your first data point.

For this situation, it would be $(0, 0)$ or $a = 0$ and $f (a) = 0$ and $(25, 5.67)$ or $b = 25$ and $f (b) = 5.67$ .

Substituting the points into the average rate of change formula results in the following:

$\frac{f (b) - f (a)}{b - a} = \frac{5.67}{25}$ , which is 0.227 miles per minute.