# Eureka Math Algebra 1 Module 3 Lesson 20 Answer Key

## Engage NY Eureka Math Algebra 1 Module 3 Lesson 20 Answer Key

### Eureka Math Algebra 1 Module 3 Lesson 20 Exploratory Challenge Answer Key

Exploratory Challenge 1
A transformation of the absolute value function f(x) = |x – 3| is rewritten here as a piecewise function. Describe in words how to graph this piecewise function.

First, I would graph the line y = – x + 3 for x – values less than 3, and then I would graph the line y = x – 3 for x – values greater than or equal to 3.

Exploratory Challenge 2
The graph y = f(x) of a piecewise function f is shown. The domain of f is – 5≤x≤5, and the range is – 1≤y≤3.
a. Mark and identify four strategic points helpful in sketching the graph of y = f(x).

( – 5, – 1), ( – 1, 1), (3, 1), and (5, 3)

b. Sketch the graph of y = 2f(x), and state the domain and range of the transformed function. How can you use part (a) to help sketch the graph of y = 2f(x)?

Domain: – 5≤x≤5, range: – 2≤y≤6. For every point (x, y) on the graph of y = f(x), there is a point (x, 2y) on the graph of y = 2f(x). The four strategic points can be used to determine the line segments in the graph of y = 2f(x) by graphing points with the same original x – coordinate and 2 times the original y – coordinate (( – 5, – 2), ( – 1, 2), (3, 2), and (5, 6)).

c. A horizontal scaling with scale factor $$\frac{1}{2}$$ of the graph of y = f(x) is the graph of y = f(2x). Sketch the graph of y = f(2x), and state the domain and range. How can you use the points identified in part (a) to help sketch y = f(2x)?

Domain: – 2.5≤x≤2.5, range: – 1≤y≤3. For every point (x, y) on the graph of y = f(x), there is a point ($$\frac{x}{2}$$, y) on the graph of y = f(2x). The four strategic points can be used to determine the line segments in the graph of y = f(2x) by graphing points with one – half the original x – coordinate and the original y – coordinate (( – 2.5, – 1), ( – 0.5, 1), (1.5, 1), and (2.5, 3)).

### Eureka Math Algebra 1 Module 3 Lesson 20 Exercise Answer Key

Opening Exercise
Fill in the blanks of the table with the appropriate heading or descriptive information.

Exercises 1–2

Exercise 1.
Describe how to graph the following piecewise function. Then, graph y = f(x) below.

The function f can be graphed of the line y = – 3x – 3 for x – values less than or equal to – 2, the graph of the line y = 0.5x + 4 for x – values greater than – 2 and less than 2, and the graph of the line y = – 2x + 9 for x – values greater than or equal to 2.

Exercise 2.
Using the graph of f below, write a formula for f as a piecewise function.

Exercises 3–4
Exercise 3.
How does the range of f in Exploratory Challenge 2 compare to the range of a transformed function g, where g(x) = kf(x), when k>1?
For every point (x, y) in the graph of y = f(x), there is a point (x, ky) in the graph of y = kf(x), where the number ky is a multiple of each y. For values of k > 1, y = kf(x) is a vertical scaling that appears to stretch the graph of y = f(x). The original range, – 1≤y≤3 for y = f(x) becomes – 1k≤y≤3k for the function y = kf(x).

Exercise 4.
How does the domain of f in Exploratory Challenge 2 compare to the domain of a transformed function g, where g(x) = f($$\frac{1}{k}$$ x), when 0<k<1? (Hint: How does a graph shrink when it is horizontally scaled by a factor k?)
For every point (x, y) in the graph of y = f(x), there is a point (kx, y) in the graph of y = f($$\frac{1}{k}$$ x). For values of 0 < k < 1, y = f($$\frac{1}{k}$$ x) is a horizontal scaling by a factor k that appears to shrink the graph of y = f(x).
This means the original domain, – 5≤x≤5 for y = f(x), becomes – 5k≤x≤5k for the function y = f($$\frac{1}{k}$$ x).

### Eureka Math Algebra 1 Module 3 Lesson 20 Problem Set Answer Key

Question 1.
Suppose the graph of f is given. Write an equation for each of the following graphs after the graph of f has been transformed as described. Note that the transformations are not cumulative.
a. Translate 5 units upward.
y = f(x) + 5

b. Translate 3 units downward.
y = f(x) – 3

c. Translate 2 units right.
y = f(x – 2)

d. Translate 4 units left.
y = f(x + 4)

e. Reflect about the x – axis.
y = – f(x)

f. Reflect about the y – axis.
y = f( – x)

g. Stretch vertically by a factor of 2.
y = 2f(x)

h. Shrink vertically by a factor of $$\frac{1}{3}$$.
y = $$\frac{1}{3}$$ f(x)

i. Shrink horizontally by a factor of $$\frac{1}{3}$$.
y = f(3x)

j. Stretch horizontally by a factor of 2.
y = f($$\frac{1}{2}$$ x)

Question 2.
Explain how the graphs of the equations below are related to the graph of y = f(x).
a. y = 5f(x)
The graph is a vertical stretch of y = f(x) by a factor of 5.

b. y = f(x – 4)
The graph of y = f(x) is translated right 4 units.

c. y = – 2f(x)
The graph is a vertical stretch of y = f(x) by a factor of 2 and reflected about the x – axis.

d. y = f(3x)
The graph is a horizontal shrink of y = f(x) by a factor of $$\frac{1}{3}$$.

e. y = 2f(x) – 5
The graph is a vertical stretch of y = f(x) by a factor of 2 and translated down 5 units.

Question 3.
The graph of the equation y = f(x) is provided below. For each of the following transformations of the graph, write a formula (in terms of f) for the function that is represented by the transformation of the graph of y = f(x). Then, draw the transformed graph of the function on the same set of axes as the graph of y = f(x).

a. A translation 3 units left and 2 units up
p(x) = f(x + 3) + 2

b. A vertical stretch by a scale factor of 3.
q(x) = 3f(x)

c. A horizontal shrink by a scale factor of $$\frac{1}{2}$$
r(x) = f(2x)

Question 4.
Reexamine your work on Exploratory Challenge 2 and Exercises 3 and 4 from this lesson. Parts (b) and (c) of Exploratory Challenge 2 asked how the equations y = 2f(x) and y = f(2x) could be graphed with the help of the strategic points found in part (a). In this problem, we investigate whether it is possible to determine the graphs of y = 2f(x) and y = f(2x) by working with the piecewise linear function f directly.
a. Write the function f in Exploratory Challenge 2 as a piecewise linear function.

b. Let g(x) = 2f(x). Use the graph you sketched in Exploratory Challenge 2, part (b) of y = 2f(x) to write the formula for the function g as a piecewise linear function.

c. Let h(x) = f(2x). Use the graph you sketched in Exploratory Challenge 2, part (c) of y = f(2x) to write the formula for the function h as a piecewise linear function.

d. Compare the piecewise linear functions g and h to the piecewise linear function f. Did the expressions defining each piece change? If so, how? Did the domains of each piece change? If so how?
Function g: Each piece of the formula for g is 2 times the corresponding piece of the formula for f.
The domains are the same.

Function h: Each piece of the formula for h is found by substituting 2x in for x in the corresponding piece of the formula for f. The length of each interval in the domain of h is $$\frac{1}{2}$$ the length of the corresponding interval in the domain of f.

### Eureka Math Algebra 1 Module 3 Lesson 20 Exit Ticket Answer Key

Question 1.
The graph of a piecewise function f is shown below.
Let p(x) = f(x – 2), q(x) = $$\frac{1}{2}$$ f(x – 2), and r(x) = $$\frac{1}{2}$$ f(x – 2) + 3.
Graph y = p(x), y = q(x), and y = r(x) on the same set of axes as the graph of y = f(x).