## Engage NY Eureka Math Algebra 1 Module 4 Lesson 18 Answer Key

### Eureka Math Algebra 1 Module 4 Lesson 18 Exploratory Challenge Answer Key

Exploratory Challenge 1

Use your graphing calculator to create a data table for the functions y = x^{2} and y = \(\sqrt{x}\) for a variety of x – values. Use both negative and positive numbers, and round decimal answers to the nearest hundredth.

Answer:

Exploratory Challenge 2

Create the graphs of y = x^{2} and y = \(\sqrt{x}\) on the same set of axes.

Answer:

Exploratory Challenge 3

Create a data table for y = x^{3} and y = \(\sqrt [ 3 ]{ x }\), and graph both functions on the same set of axes. Round decimal answers to the nearest hundredth.

Answer:

### Eureka Math Algebra 1 Module 4 Lesson 18 Exercise Answer Key

Opening Exercise

a. Evaluate x^{2} when x = 7.

Answer:

49

b. Evaluate \(\sqrt{x}\) when x = 81.

Answer:

9

c. Evaluate x^{3} when x = 5.

Answer:

125

d. Evaluate \(\sqrt [ 3 ]{ x }\) when x = 27.

Answer:

3

### Eureka Math Algebra 1 Module 4 Lesson 18 Problem Set Answer Key

Question 1.

Create the graphs of the functions f(x) = x^{2} + 2 and g(x) = \(\sqrt{x}\) + 2 using the given values. Use a calculator to help with decimal approximations.

Answer:

See the values in the table.

Question 2.

What can be said about the first three values for g(x) in the table?

Answer:

The domain of g(x) = \(\sqrt{x}\) + 2 is limited to nonnegative numbers since the square root of a negative number is not real, so there is no value for g( – 4), g( – 2), or g( – 1).

Question 3.

Describe the relationship between the graphs given by the equations y = x^{2} + 2 and y = \(\sqrt{x}\) + 2. How are they alike? How are they different?

Answer:

The graph of the square root function is a reflection of the graph of the quadratic function when x is nonnegative. The reflection is about the line given by the graph of the equation y = x + 2. The domain of y = x^{2} + 2 is all real numbers. The domain of y = \(\sqrt{x}\) + 2 is x≥0. The range of y = x^{2} + 2 and y = \(\sqrt{x}\) + 2 is y≥2.

Question 4.

Refer to your class notes for the graphs of y = x^{2} and y = \(\sqrt{x}\). How are the graphs of y = x^{2} and y = \(\sqrt{x}\) transformed to generate the graphs of y = x^{2} + 2 and y = \(\sqrt{x}\) + 2?

Answer:

The graph of y = x^{2} + 2 is the graph of y = x^{2} translated vertically 2 units up. The graph of y = \(\sqrt{x}\) + 2 is also the vertical translation of the graph of y = \(\sqrt{x}\) translated 2 units up.

Question 5.

Create the graphs of p(x) = x^{3} – 2 and q(x) = \(\sqrt [ 3 ]{ x }\) – 2 using the given values for x. Use a calculator to help with decimal approximations.

Answer:

Question 6.

For the table in Problem 5, explain why there were no function values that resulted in an error.

Answer:

Unlike square roots, the domain of a cube root function includes all real numbers since the product of three (or any other odd number) factors of a negative number, yields a negative number. Since the domains for both functions include all real numbers, there are no excluded rows in the table.

Question 7.

Describe the relationship between the domains and ranges of the functions p(x) = x^{3} – 2 and q(x) = \(\sqrt [ 3 ]{ x }\) – 2. Describe the relationship between their graphs.

Answer:

The domain and range of p(x) = x^{3} – 2 and q(x) = \(\sqrt [ 3 ]{ x }\) – 2 are all real numbers. The graphs of y = x^{3} – 2 and y = \(\sqrt [ 3 ]{ x }\) – 2 are each symmetrical about the line given by the equation y = x – 2.

Question 8.

Refer to your class notes for the graphs of y = x^{3} and y = \(\sqrt [ 3 ]{ x }\). How are the graphs of y = x^{3} and y = \(\sqrt [ 3 ]{ x }\) transformed to generate the graphs of y = x^{3} – 2 and y = \(\sqrt [ 3 ]{ x }\) – 2?

Answer:

The graph of y = x^{3} – 2 is the graph of y = x^{3} translated vertically 2 units down. The graph of y = \(\sqrt [ 3 ]{ x }\) – 2 is also the vertical translation of the graph of y = \(\sqrt [ 3 ]{ x }\) translated 2 units down.

Question 9.

Using your responses to Problems 4 and 8, how do the functions given in Problems 1 and 5 differ from their parent functions? What effect does that difference seem to have on the graphs of those functions?

Answer:

In Problem 1, f is the squaring function x^{2} plus 2, and g is the square root function \(\sqrt{x}\) plus 2. Adding 2 to a function translates the graph of the function 2 units up vertically. In Problem 5, p is the cubing function x^{3} minus 2, and q is the cube root function \(\sqrt [ 3 ]{ x }\) minus 2. Subtracting 2 from a function translates the graph of the function 2 units down vertically.

Question 10.

Create your own functions using r(x) = x^{2} – ☐ and s(x) = \(\sqrt{x}\) – ☐ by filling in the box with a positive or negative number. Predict how the graphs of your functions will compare to the graphs of their parent functions based on the number that you put in the blank boxes. Generate a table of solutions for your functions, and graph the solutions.

Answer:

Answers will vary. If k is the number inserted into ☐, then the graph of the function will be translated vertically k units down for positive k values and – k units up for negative k values.

### Eureka Math Algebra 1 Module 4 Lesson 18 Exit Ticket Answer Key

Question 1.

Describe the relationship between the graphs of y = x^{2} and y = \(\sqrt{x}\). How are they alike? How are they different?

Answer:

The square root function is a reflection of the quadratic function about y = x when x is nonnegative. The domain of y = x^{2} is all real numbers. The domain of y = \(\sqrt{x}\) is x≥0. The range of y = x^{2} and y = \(\sqrt{x}\) is y≥0.

Question 2.

Describe the relationship between the graphs of y = x^{3} and y = \(\sqrt [ 3 ]{ x }\). How are they alike? How are they different?

Answer:

The domain and range of y = x^{3} and y = \(\sqrt [ 3 ]{ x }\) are all real numbers. The shape of the graphs of y = x^{3} and y = \(\sqrt [ 3 ]{ x }\) are the same, but they are oriented differently. (Some students may be able to articulate that each graph appears to be a reflection of the other across a diagonal line going through the origin.)