## Engage NY Eureka Math Algebra 1 Module 3 Lesson 17 Answer Key

### Eureka Math Algebra 1 Module 3 Lesson 17 Exploratory Challenge Answer Key

Exploratory Challenge 1

Let f(x) = |x|, g(x) = f(x) – 3, and h(x) = f(x) + 2 for any real number x.

a. Write an explicit formula for g(x) in terms of |x| (i.e., without using f(x) notation).

Answer:

g(x) = |x| – 3

b. Write an explicit formula for h(x) in terms of |x| (i.e., without using f(x) notation).

Answer:

h(x) = |x| + 2

c. Complete the table of values for these functions.

Answer:

d. Graph all three equations: y = f(x), y = f(x) – 3, and y = f(x) + 2.

Answer:

e. What is the relationship between the graph of y = f(x) and the graph of y = f(x) + k?

Answer:

For values of k where k>0, for every point (x, f(x)) that satisfies the equation y = f(x), there is a corresponding point (x, f(x) + k) on the graph, located k units above (x, f(x)), that satisfies the equation y = f(x) + k. The graph of y = f(x) + k is the vertical translation of the graph of y = f(x) by k units upward.

For values of k where k<0, for every point (x, f(x)) that satisfies the equation y = f(x), there is a corresponding point (x, f(x) + k) on the graph, located k units below (x, f(x)), that satisfies the equation y = f(x) + k. The graph of y = f(x) + k is the vertical translation of the graph of y = f(x) by k units downward.

f. How do the values of g and h relate to the values of f?

Answer:

For each x in the domain of f and g, the value of g(x) is 3 less than the value of f(x). For each x in the domain of f and h, the value of h(x) is 2 more than the value of f(x).

Exploratory Challenge 2

Let f(x) = |x|, g(x) = 2f(x), and h(x) = \(\frac{1}{2}\) f(x) for any real number x.

a. Write a formula for g(x) in terms of |x| (i.e., without using f(x) notation).

Answer:

g(x) = 2|x|

b. Write a formula for h(x) in terms of |x| (i.e., without using f(x) notation).

Answer:

h(x) = \(\frac{1}{2}\)|x|

c. Complete the table of values for these functions.

Answer:

d. Graph all three equations: y = f(x), y = 2f(x), and y = \(\frac{1}{2}\) f(x).

Answer:

Given f(x) = |x|, let p(x) = – |x|, q(x) = – 2f(x), and r(x) = – \(\frac{1}{2}\) f(x) for any real number x.

e. Write the formula for q(x) in terms of |x| (i.e., without using f(x) notation).

Answer:

q(x) = – 2|x|

f. Write the formula for r(x) in terms of |x| (i.e., without using f(x) notation).

Answer:

r(x) = – \(\frac{1}{2}\) |x|

g. Complete the table of values for the functions p(x) = – |x|, q(x) = – 2f(x), and r(x) = – \(\frac{1}{2}\) f(x).

Answer:

h. Graph all three functions on the same graph that was created in part (d). Label the graphs as y = p(x), y = q(x), and y = r(x).

Answer:

i. How is the graph of y = f(x) related to the graph of y = kf(x) when k > 1?

Answer:

The graph of y = kf(x) for k > 1 contains points (x, ky), which are related to points (x, y) in the graph of y = f(x). The number ky is a multiple of y: Each y – value of y = g(x) is k times the y – value of y = f(x). The graph of y = kf(x) is a vertical scaling that appears to stretch the graph of y = f(x) vertically by a factor of k.

j. How is the graph of y = f(x) related to the graph of y = kf(x) when 0 < k < 1?

Answer:

The graph of y = kf(x) for 0<k<1 contains points (x, ky), which are related to points (x, y) in the graph of y = f(x). The number ky is a fraction of y: Each y – value of y = g(x) is k times the y – value of y = f(x). The graph of y = kf(x) is a vertical scaling that appears to shrink the graph of y = f(x) vertically by a factor of k.

k. How do the values of functions p, q, and r relate to the values of functions f, g, and h, respectively?

What transformation of the graphs of f, g, and h represents this relationship?

Answer:

Each function is the opposite of the corresponding function. The result is that each y – value of any point on the graph of y = p(x), y = q(x), and y = r(x) are the opposite of the y – value of the graphs of the equations

y = f(x), y = g(x), and y = h(x). Each graph is a reflection of the corresponding graph over the x – axis.

### Eureka Math Algebra 1 Module 3 Lesson 17 Exercise Answer Key

Exercise

Make up your own function f by drawing the graph of it on the Cartesian plane below. Label it as the graph of the equation y = f(x). If b(x) = f(x) – 4 and c(x) = 1/4 f(x) for every real number x, graph the equations y = b(x) and y = c(x) on the same Cartesian plane.

Answer:

Answers will vary. Look for and encourage students to create interesting graphs for their function f. (Functions DO NOT have to be defined by algebraic expressionsâ€”any graph that satisfies the definition of a function will do.) One such option is using f(x) = |x| as shown in the example below.

### Eureka Math Algebra 1 Module 3 Lesson 17 Problem Set Answer Key

Let f(x) = |x| for every real number x. The graph of y = f(x) is shown below. Describe how the graph for each function below is a transformation of the graph of y = f(x). Then, use this same set of axes to graph each function for Problems 1â€“5. Be sure to label each function on your graph (by y = a(x), y = b(x), etc.).

Question 1.

a(x) = |x| + \(\frac{3}{2}\)

Answer:

Translate the graph of y = f(x) up 1.5 units.

Question 2.

b(x) = – |x|

Answer:

Reflect y = f(x) across the x – axis.

Question 3.

c(x) = 2|x|

Answer:

Vertically scale/stretch the graph of y = f(x) by doubling the output values for every input.

Question 4.

d(x) = \(\frac{1}{3}\)|x|

Answer:

Vertically scale/shrink the graph of y = f(x) by dividing the output values by 3 for every input.

Question 5.

e(x) = |x| – 3

Answer:

Translate the graph of y = f(x) down 3 units.

Question 6.

Let r(x) = |x| and t(x) = – 2|x| + 1 for every real number x. The graph of y = r(x) is shown below. Complete the table below to generate output values for the function t, and then graph the equation y = t(x) on the same set of axes as the graph of y = r(x).

Answer:

Question 7.

Let f(x) = |x| for every real number x. Let m and n be functions found by transforming the graph of y = f(x). Use the graphs of y = f(x), y = m(x), and y = n(x) below to write the functions m and n in terms of the function f. (Hint: What is the k?)

Answer:

m(x) = 2f(x)

n(x) = f(x) + 2

### Eureka Math Algebra 1 Module 3 Lesson 17 Exit Ticket Answer Key

Let p(x) = |x| for every real number x. The graph of y = p(x) is shown below.

Question 1.

Let q(x) = – \(\frac{1}{2}\) |x| for every real number x. Describe how to obtain the graph of y = q(x) from the graph of y = p(x). Sketch the graph of y = q(x) on the same set of axes as the graph of y = p(x).

Answer:

Reflect and vertically scale the graph of y = p(x) by plotting (x, – \(\frac{1}{2}\) y) for each point (x, y) in the graph of y = p(x). See the graph of y = q(x) below.

Question 2.

Let r(x) = |x| – 1 for every real number x. Describe how to obtain the graph of y = r(x) from the graph of y = p(x). Sketch the graph of y = r(x) on the same set of axes as the graphs of y = p(x) and y = q(x).

Answer:

Translate the graph of y = p(x) vertically down 1 unit. See the graph of y = r(x) below.