Key Concepts
- Graph the absolute value function
- Transform the absolute value function
- Interpret the graph of a function
- Determine rate of change
Graph the Absolute Value Function
What are the features of the graph of f(x) = |x|?
Make a table of values and graph the absolute value function f(x) = |x|.
The graph has an axis of symmetry, which intersects the vertex and divides the graph into two sections, or pieces, that are images of each other under a reflection.
Transform the Absolute Value Function
A. How do domain and range of g(x) = 2 |x| compare with the domain and range of
f(x) = |x|?
Compare the graphs of g and f.
The domain of g and f are all real numbers.
Because the absolute value expression produces only non-negative values, the range of f is
y ≥ 0.
Multiplying IxI by a positive factor, in this case, 2, yields non-negative outputs, so the range of g is also y≥ 0.
The domain and range of the function g are the same as those of function f.
B. How do domain and range of h(x) = –1|x| compare with the domain and range of
f(x) = |x|?
Compare the graphs of h and f.
The domain of h and the domain of f is all real numbers.
The range of f is: y ≥ 0.
Multiplying IxI by a negative factor, -1, yields non-positive outputs, so the range of h is also y ≤ 0.
The domain of h is the same as the domain of f. The range of h is the opposite of the range
of f.
Interpret the Graph of a Function
Jay rides in a boat from Monroe County to Collier County. The graph of the function
d(t) = 30 |t-1.5| shows the distance of the boat in miles from the county line at t hours. Assume, the graph shows Jay’s entire trip.
- How far does Jay travel to visit his friend?
Jay began his trip 45 mi from the county line, traveled towards the county line, which he crossed after an hour and a half. He then traveled away from the county line and was 45 mi from the county line after 3 h. He traveled a total of 90 mi.
- How does the graph relate to the domain and range of the function?
Since Jay’s entire trip is 3 h, the domain of the function is: 0 ≤ t ≤ 3.
For 1.5 < t < 3, the section of the domain 0 < t < 1.5, his distance to the border is decreasing. For, his distance from the border is increasing.
The maximum and minimum values on the graph are 45 and 0, so the range of the function is 0 ≤ d(t) ≤ 45.
Determine Rate of Change
The graph shows Jay’s boat ride across the state line from the previous example. What is the rate of change over the interval 2 ≤ t ≤ 2.5 What does it mean in terms of the situation?
Use the slope formula to determine the rate of change from t = 2 to t = 2.5.
Use the points from (2, 15) and (2.5, 30).
The rate of change over this interval is 30.
The rate of change represents the speed of the boat in miles per hour. Since the rate of change is positive, Jay’s distance from the border is increasing. The boat is traveling at 30 mi/h away from the border.
Questions
Question 1
What are the domain and range of f(x) = |x|?
Solution:
Below is the graph of f(x) = |x|.
As we can see in the graph, the value of x can be anything. So, the domain of f is all real numbers.
In the graph, the value of f(x) is always positive or zero. So, the range of the function is y ≥ 0.
Question 2
A cyclist competing in a race ride past a water station. The graph of the function d(t) = (1/3) |t – 60| shows her distance from the water station at t minutes. Assume, the graph represents the entire race. What does the graph tell you about her race?
Solution:
The cyclist traveled 20 km towards the water station in one hour and at the end of one hour, she reached the water station. Then she started traveling away from the water station and covered the distance of 20 km away from the water station in the next one hour.
So, the cyclist finished her entire race of 40 km in two hours.
Question 3
Kata gets on a moving walkway to the airport. The, 8 s after she gets on, she taps Lisa, who is standing alongside the walkway. The graph shows Kata’s distance from Lisa over time. Calculate the rate of change in her distance from Lisa from 6 s to 8 s, and then from 8 s to
12 s. What do the rates of change mean in terms of Kata’s movement?
Solution:
Let’s find the value of d(t) at t = 6 s, t = 8 s and t = 12 s.
t = 6, d(t) = 4
t = 8, d(t) = 0
t = 12, d(t) = 10
Rate of change from 6 s to 8 s:
Use the points (6, 4) and (8, 0).
m = (y2 – y1)/ (x2 – x1) = (0 – 4)/ (8 – 6) = (-4)/2 = -2 ft/s
Rate of change from 8 s to 12 s:
Use the points (8, 0) and (12, 10).
m = (y2 – y1)/ (x2 – x1) = (10 – 0)/ (12 – 8) = 10/4 = 2.5 ft/s
Since the rate of change from 6 s to 8 s is negative, Kata’s distance from Lisa is decreasing and she is traveling at the speed of 2 ft/s toward Lisa. And then from 8 s to 12 s, the rate of change is positive. So, Kata’s distance from Lisa is increasing, and is traveling at the speed of 2.5 ft/s moving away from Lisa.
Key Concepts Covered
Exercise
Graph the following functions:
- f(x) = 0.6|x|
- f(x) = (1/7) |x|
- f(x) = |x – 100|
- f(x) = 3|x|
- f(x) = 2Ix – 100I
- f(x) = (1/3) |x – 100|
- f(x) = 4.5|x|
- f(x) = 0.5|x + 10|
- f(x) = 10|x|
- f(x) = 9|x + 100|
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