TBIL-Precal Slope and Average Rate of Change (LF1)

🔗

Section 3.1 Slope and Average Rate of Change (LF1)

🔗

Objectives

Determine the average rate of change of a given function over a given interval. Find the slope of a line.

🔗

Subsection 3.1.1 Activities

🔗

Remark 3.1.1.

This section will explore ideas around average rate of change and slope. To help us get started, let’s take a look at a context in which these ideas can be helpful.

🔗

Activity 3.1.2.

🔗

Robert came home one day after school to a very hot house! When he got home, the temperature on the thermostat indicated that it was

85

degrees! Robert decided that was too hot for him, so he turned on the air conditioner. The table of values below indicate the temperature of his house after turning on the air conditioner.

Table 3.1.3.

Time (minutes)	Temperature (degrees Fahrenheit)
$0$	$85$
$1$	$84.3$
$2$	$83.6$
$3$	$82.9$
$4$	$82.2$
$5$	$81.5$
$6$	$80.8$

🔗

(a)

🔗

How much did the temperature change from

0

2

minutes?

The temperature decreased by $0.7$ degrees
The temperature decreased by $1.4$ degrees
The temperature increased by $0.7$ degrees
The temperature increased by $1.4$ degrees

Answer.

🔗

(b)

🔗

How much did the temperature change from

4

6

minutes?

The temperature decreased by $0.7$ degrees
The temperature decreased by $1.4$ degrees
The temperature increased by $0.7$ degrees
The temperature increased by $1.4$ degrees

Answer.

🔗

(c)

🔗

If Robert wanted to know how much the temperature was decreasing each minute, how could he figure that out?

Answer.

Choosing any two points that have a time difference of

1

minute should show that the temperature decreases by

0.7

degrees every minute.

🔗

(d)

🔗

How would you describe the overall behavior of the temperature of Robert’s house?

Answer.

Overall, the temperature of Robert’s house is decreasing (the air conditioner appears to be working!). Notice that in parts (a) and (b), the temperature decreased

1.4

degrees in

2

minutes.

🔗

Remark 3.1.4.

Notice in Activity 3.1.2 that the temperature appears to be decreasing at a constant rate (i.e., the temperature decreased

1.4

degrees for every

2

-minute interval). Upon further investigation, you might have also noticed that the temperature decreased by

0.7

degrees every minute.

🔗

Activity 3.1.5.

🔗

Refer back to the data Robert collected of the temperature of his house after turning on the air conditioner (Table 3.1.3).

🔗

(a)

🔗

If this pattern continues, what will the temperature be after

8

minutes?

$80.1$
$78.7$
$80.8$
$79.4$

Answer.

🔗

(b)

🔗

If this pattern continues, how long will it take for Robert’s house to reach

78

degrees?

$12$ minutes
$9$ minutes
$10$ minutes
$11$ minutes

Answer.

🔗

Remark 3.1.6.

An average rate of change helps us to see and understand how a function is generally behaving. For example, in Activity 3.1.2 and Activity 3.1.5, we began to see how the temperature of Robert’s house was decreasing every minute the air conditioner was on. In other words, when looking at average rate of change, we are comparing how one quantity is changing with respect to something else changing.

🔗

Definition 3.1.7.

🔗

An average rate of change of a function calculates the amount of change in one item divided by the corresponding amount of change in another.

🔗

To calculate the average rate of change for any function

f (x),

we pick two points,

a

and

b,

and evaluate the function at those two points. We then find the difference between the

y

-values and

x

-values to calcuate the average rate of change:

\frac{f (b) - f (a)}{b - a} .

🔗

Activity 3.1.8.

🔗

Use the table below to answer the questions.

Table 3.1.9.

$x$	$f (x)$
$- 5$	$28$
$- 4$	$19$
$- 3$	$12$
$- 2$	$7$
$- 1$	$4$

🔗

(a)

🔗

Applying Definition 3.1.7, what is the average rate of change when

x = - 5

x = - 2 ?

$\frac{1}{7}$
$- 3$
$- 7$
$7$

Answer.

🔗

(b)

🔗

What is the average rate of change on the interval

[- 4, - 1] ?

$- 5$
$- 3$
$5$
$3$

Answer.

🔗

(c)

🔗

Does this function have a constant average rate of change?

Answer.

No. This would be a great opportunity for students to make observations of the average rate of change across different intervals.

🔗

Activity 3.1.10.

🔗

Use the graph to calculate the average rate of change on the given intervals.

🔗

(a)

🔗

Applying Definition 3.1.7, what is the average rate of change on the interval

[- 4, 0] ?

$- \frac{1}{2}$
$\frac{1}{2}$
$- 2$
$2$

Answer.

🔗

(b)

🔗

What is the average rate of change on the interval

[0, 12] ?

$\frac{1}{6}$
$- 6$
$- \frac{1}{6}$
$6$

Answer.

🔗

Activity 3.1.11.

🔗

Just like with tables and graphs, you should be able to find the average rate of change when given a function. For this activity, use the function

f (x) = - 3 x^{2} - 1

🔗

to answer the following questions.

🔗

(a)

🔗

Applying Definition 3.1.7, what is the average rate of change on the interval

[- 2, 3] ?

$\frac{41}{5}$
$- 3$
$- \frac{1}{3}$
$\frac{5}{41}$

Answer.

🔗

(b)

🔗

What is the average rate of change on the interval

[0, 4] ?

$\frac{2}{25}$
$- 50$
$- \frac{1}{12}$
$- 12$

Answer.

🔗

Activity 3.1.12.

🔗

Use the given graph of the function,

f (x) = 3 x - 4,

to investigate the average rate of change of a linear function.

🔗

(a)

🔗

What is the average rate of change on the interval

[- 2, 0] ?

$\frac{1}{3}$
$3$
$- \frac{1}{7}$
$- 7$

Answer.

🔗

(b)

🔗

What is the average rate of change on the interval

[- 1, 5] ?

Notice that you cannot see the point at

x = 5 .

How could you use the equation of the line to determine the

y

-value when

x = 5 ?

$3$
$\frac{1}{3}$
$\frac{2}{3}$
$- 3$

Answer.

🔗

(c)

🔗

Based on your observations in parts (a) and (b), what do you think will be the average rate of change on the interval

[5, 25] ?

Answer.

The average rate of change is

3 .

Based on the previous parts, it shows that the average rate of change does not change for this function.

🔗

Remark 3.1.13.

Notice in Activity 3.1.12, the average rate of change was the same regardless of which interval you were given. But in Activity 3.1.10, the average rate of change was not the same across different intervals.

🔗

Definition 3.1.14.

🔗

The slope of a line has a constant that represents the direction and steepness of the line. For a linear function, the slope never changes - meaning it has a constant average rate of change.

🔗

Activity 3.1.15.

🔗

The steepness of a line depends on the vertical and horizontal distances between two points on the line. Use the graph below to compare the steepness, or slope, of the two lines.

🔗

(a)

🔗

What is the vertical distance between the two points on the line

y = g (x)

(the red line)?

$2$
$4$
$8$
$\frac{1}{2}$

Answer.

🔗

(b)

🔗

What is the horizontal distance between the two points on the line

y = g (x)

(the red line)?

$2$
$4$
$8$
$\frac{1}{2}$

Answer.

🔗

(c)

🔗

Using information from parts (a) and (b), what value could we use to describe the steepness of the line

y = g (x)

(the red line)?

$2$
$4$
$8$
$\frac{1}{2}$

Answer.

B. This could be a good place to discuss where

4

comes from (change of

y

over the change in

x

🔗

(d)

🔗

What is the vertical distance between the two points on the line

y = f (x)

(the blue line)?

$2$
$4$
$8$
$\frac{1}{2}$

Answer.

🔗

(e)

🔗

What is the horizontal distance between the two points on the line

y = f (x)

(the blue line)?

$2$
$4$
$8$
$\frac{1}{2}$

Answer.

🔗

(f)

🔗

Using information from parts (d) and (e), what value could we use to describe the steepness of the line

y = f (x)

(the blue line)?

$2$
$4$
$8$
$\frac{1}{2}$

Answer.

D. This could be a good place to discuss where

\frac{1}{2}

comes from (change of

y

over the change in

x

🔗

(g)

🔗

Which line is the steepest?

Answer.

The line

y = g (x)

(the red line). The slope of

y = g (x)

4

and the slope of

y = f (x)

(the blue line) is

\frac{1}{2} .

Because

4

is larger than

\frac{1}{2},

the line

y = g (x)

(the red line) is steeper.

🔗

Remark 3.1.16.

The steepness, or slope, of a line can be found by the change in

y

(the vertical distance between two points on the line) divided by the change in

x

(the horizontal distance between two points on the line). Slope can be calculated as "rise over run."

🔗

Slope is a way to describe the steepness of a line. The red line in Activity 3.1.15 has a larger value for it’s slope than the blue line. Thus, the red line is steeper than the blue line.

🔗

Activity 3.1.17.

🔗

Now that we know how to find the slope (or steepness) of a line, let’s look at other properties of slope. Use the graph below to answer the following questions.

🔗

(a)

🔗

What is the slope of the line

y = f (x)

(the blue line)?

$\frac{1}{2}$
$2$
$- \frac{1}{2}$
$- 2$

Answer.

🔗

(b)

🔗

What is the slope of

y = g (x)

(the green line)?

$\frac{1}{2}$
$2$
$- \frac{1}{2}$
$- 2$

Answer.

🔗

(c)

🔗

How are the slopes of the lines similar?

Answer.

Students will probably notice that both slopes have the same steepness

(\frac{1}{2}) .

🔗

(d)

🔗

How are the slopes of the lines different?

Answer.

Although both slopes have the same steepness

(\frac{1}{2}),

one line has a positive slope (

y = f (x)

, the blue line) and the other line has a negative slope (

y = g (x),

the green line).

🔗

Remark 3.1.18.

Notice in Activity 3.1.17 that the slope does not just indicate how steep a line is, but also it’s direction. A negative slope indicates that the line is decreasing (from left to right) and a positive slope indicates that the line is increasing (from left to right).

🔗

Activity 3.1.19.

🔗

Suppose

(- 3, 7)

and

(7, 2)

are two points on a line.

🔗

(a)

🔗

Plot these points on a graph and find the slope by using "rise over run."

$\frac{1}{2}$
$2$
$- \frac{1}{2}$
$- 2$

Answer.

🔗

(b)

🔗

Now calculate the slope by using the change in

y

over the change in

x .

$\frac{1}{2}$
$2$
$- \frac{1}{2}$
$- 2$

Answer.

🔗

(c)

🔗

What do you notice about the slopes you got in parts (a) and (b)?

Answer.

Students should notice that they get the same answer. Both methods can be used to find the slope.

🔗

Remark 3.1.20.

We can calculate slope (

m

) by finding the change in

y

and dividing by the change in

x .

Mathematically, this means that when given

(x_{1}, y_{1})

and

(x_{2}, y_{2}),

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} .

🔗

Activity 3.1.21.

🔗

Calculate the slope of each representation of a line using the slope formula.

🔗

(a)

🔗

$x$	$f (x)$
$- 2$	$- 7$
$- 1$	$- 4$
$0$	$- 1$
$1$	$2$
$2$	$5$

Answer.

m = 3

🔗

(b)

Answer.

m = 0

🔗

(c)

🔗

(- 4, 7)

and

(- 4, 1)

Answer.

The slope is undefined.

🔗

Remark 3.1.22.

In Activity 3.1.21, there were slopes that were

0

and undefined. When a line is vertical, the slope is undefined. This means that there is only a vertical distance between two points and there is no horizontal distance. When a line is horizontal, the slope is

0 .

This means that the line never rises vertically, giving a vertical distance of zero.

🔗

Exercises 3.1.2 Exercises

🔗

Exercises available at https://tbil.org/precalculus/preview/exercises/#/bank/LF1/.

Prev Top Next