## Engage NY Eureka Math Algebra 2 Module 3 Lesson 20 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 20 Opening Exercise Answer Key

Opening Exercise:

a. Sketch the graphs of the three functions f(x) = x^{2}, g(x) = (2x)^{2} + 1, and h(x) = 4x^{2} + 1.

i. Describe the sequence of transformations that takes the graph of f(x) = x^{2} to the graph of g(x) = (2x)^{2} + 1.

Answer:

The graph of g is a horizontal scaling by a factor of \(\frac{1}{2}\) and a vertical translation up 1 unit of the graph of f.

ii. Describe the sequence of transformations that takes the graph of f(x) = x^{2} to the graph of h(x) = 4x^{2} + 1.

Answer:

The graph of h is a vertical scaling by a factor of 4 and a vertical translation up 1 unit of the graph of f.

iii. Explain why g and h from parts (I) and (ii) are equivalent functions.

Answer:

These functions are equivalent and have the same graph because the expressions (2x)^{2} + 1 and 4x^{2} + 1 are equivalent. The blue graph shown is the graph off, and the green graph is the graph of g and h.

b. Describe the sequence of transformations that takes the graph of f(x) = sin(x) to the graph of g(x) = sin(2x) – 3.

Answer:

The graph of g is a horizontal scaling by a factor of \(\frac{1}{2}\) and a vertical translation down 3 units of the graph of f.

c. Describe the sequence of transformations that takes the graph of f(x) = sin(x) to the graph of h(x) = 4 sin(x) –Â 3.

Answer:

The graph of h is a vertical scaling by a factor of 4 and a vertical translation down 3 units of the graph of f.

d. Explain why g and h from parts (b) – (c) are not equivalent functions.

Answer:

These functions are not equivalent because they do not have the same graphs, and the two expressions are not equivalent.

### Eureka Math Algebra 2 Module 3 Lesson 20 Exploratory Challenge Answer Key

Exploratory Challenge:

a. Sketch the graph of f(x) = log_{2}(x) by identifying and plotting at least five key points. Use the table below to get started.

x | Log_{2}(x) |

\(\frac{1}{4}\) | -2 |

\(\frac{1}{2}\) | -1 |

1 | |

2 | |

4 | |

8 |

Answer:

The graph of f is shown below.

x | Log_{2}(x) |

\(\frac{1}{4}\) | -2 |

\(\frac{1}{2}\) | -1 |

1 | 0 |

2 | 1 |

4 | 2 |

8 | 3 |

b. Describe a sequence of transformations that takes the graph off to the graph of g(x) = log2(8x).

Answer:

The graph of g is a horizontal scaling by a factor of \(\frac{1}{8}\) of the graph of f.

c. Describe a sequence of transformations that takes the graph off to the graph of h(x) = 3 + log2(x).

Answer:

The graph of h is a vertical translation up 3 units of the graph of f.

d. Complete the table below for f, g, and h and describe any noticeable patterns.

Answer:

The functions g and h have the same range values at each domain value in the table.

e. Graph the three functions on the same coordinate axes and describe any noticeable patterns.

Answer:

The graphs of g and h are identical.

f. Use a property of logarithms to show that g and h are equivalent.

Answer:

By the product property and the definition of logarithm, log_{2}(8x) = log_{2}(8) + log_{2}(x) = 3 + log_{2}(x), so g(x) = h(x) for all positive real numbers x and g and h are equivalent functions.

g. Describe the graph of p(x) = log_{2}(\(\frac{x}{4}\)) as a vertical translation of the graph of f(x) = log2(x). Justify your response.

Answer:

The graph of p isa vertical translation down 2 units of the graph off because log_{2}(\(\frac{x}{4}\)) = log_{2}(x) – 2.

h. Describe the graph of h(x) = 3 + log_{2}(x) as a horizontal scaling of the graph of f(x) = log_{2}(x). Justify your response.

Answer:

The graph of h is a horizontal scaling by a factor of \(\frac{1}{8}\) of the graph of f because

3 + log_{2}(x)= log_{2}(8) + log_{2}(x)= log_{2}(8x).

i. Do the functions h(x) = log_{2}(8) + log_{2}(x) and k(x) = log_{2}(x + 8) have the same graphs? Justify your reasoning.

Answer:

No, they do not. By substituting 1 for x in both f and g, you can see that the graphs of the two functions do not have the same y-coordinate at this point. Therefore, the graphs cannot be the same if at least one point is different.

j. Use properties of exponents to explain why graphs of f(x) = 4^{x} and g(x) = (2^{2})^{x} are identical.

Answer:

Using the power property of exponents, 2^{2x} = (2^{2})x = 4^{x} Since the expressions are equal, the graphs of the functions would be the same.

k. Use the properties of exponents to predict what the graphs of g(x) = 4 . 2^{x} and h(x) = 2^{x + 2} look like compared to one another. Describe the graphs of g and h as transformations of the graph of f(x) = 2^{x}. Confirm your prediction by graphing f, g, and h on the same coordinate axes.

Answer:

The graphs of the two functions g and h are the same since 2^{x + 2} = 2^{x} . 2^{2} = 4. 2^{x} by the multiplication property of exponents and the commutative property. The graph of g is the graph of f(x) = 2^{x} scaled vertically by a factor of 4. The graph of h is the graph of f(x) = 2^{x} translated horizontally 2 units to the left.

l. Graph f(x) = 2^{x}, g(x) = 2^{-x} and h(x) = (\(\frac{1}{2}\))^{x} on the same coordinate axes. Describe the graphs of g and h as transformations of the graph off. Use the properties of exponents to explain why g and h are equivalent.

Answer:

The graph of g and the graph of h are both reflections about the vertical axis of the graph of f. They are equivalent because ((\(\frac{1}{2}\)))^{x} = (2^{-1})^{-x} = 2^{-x} by the definition of a negative exponent and the power property of exponents.

### Eureka Math Algebra 2 Module 3 Lesson 20 Example Answer Key

Example 1: Graphing Transformations of the Logarithm Functions

The general form of a logarithm function is given by f(x) = k + a log(x – h), where a, b, k, and h are real numbers such that b is a positive number not equal to 1, and x – h > 0.

a. Given g(x) = 3 + 2 log(x – 2), describe the graph of g as a transformation of the common logarithm function.

Answer:

The graph of g is a horizontal translation 2 units to the right, a vertical scaling by a factor of 2, and a vertical translation up 3 units of the graph of the common logarithm function.

b. Graph the common logarithm function and g on the same coordinate axes.

Answer:

Example 2: Graphing Transformations of Exponential Functions

The general form of the exponential function is given by f(x) = a . b^{x} + k, where a, b, and k are real numbers such that b is a positive number not equal to 1.

a. Use the properties of exponents to transform the function g(x) = 3^{2x + 1} – 2 to the general form, and then graph it. What are the values of a, b, and k?

Answer:

Using the properties of exponents, 3^{2x + 1} – 2 = 3^{2x} . 3^{1} – 2 = 3 . 9^{x} – 2. Thus, g(x) = 3(9)^{x} – 2 so a = 3, b = 9, and k = -2.

b. Describe the graph of g as a transformation of the graph of f(x) = 9^{x}.

Answer:

The graph of g is a vertical scaling by a factor of 3 and a vertical translation down 2 units of the graph of f.

c. Describe the graph of g as a transformation of the graph of f(x) = 3^{x}.

Answer:

The graph of g is a horizontal scaling by a factor of \(\frac{1}{2}\), a vertical scaling by a factor of 3, and a vertical translation down 2 units of the graph of f.

d. Sketch the graph of g using transformations.

Answer:

The graph of f(x) = 9^{x} is shown in blue, and the graph of g is shown in green.

### Eureka Math Algebra 2 Module 3 Lesson 20 Exercise Answer Key

Exercises:

Graph each pair of functions by first graphing f and then graphing g by applying transformations of the graph of f. Describe the graph of g as a transformation of the graph of f.

Exercise 1.

f(x) = log_{3}(x) and g(x) = 2 log_{3}(x – 1)

Answer:

The graph of g is the graph of f translated 1 unit to the right and stretched vertically by a factor of 2.

Exercise 2.

f(x) = log(x) and g(x) = log(100x)

Answer:

Because of the product property of logarithms, g(x) = 2 + log(x). The graph of g is the graph of f translated vertically 2 units.

Exercise 3.

f(x) = log_{5}x and g(x) = – log_{5}(5(x + 2))

Answer:

Since -log_{5}(5(x + 2)) = -1 – log_{5}(x + 2) by the product property of logarithms and the distributive property, the graph of gis the graph off translated 2 units to the left, reflected across the horizontal axis, and translated down 1 unit.

Exercise 4.

f(x) = 3^{x} and g(x) = -2 . 3^{x – 1}

Answer:

Since -2 . 3^{x – 1} = -2 . 3^{x}. 3^{x – 1} = –\(\frac{2}{3}\) . 3^{x} by the properties of exponents and the commutative property, the graph of g is the graph of f reflected across the horizontal axis and compressed by afactor of \(\frac{2}{3}\).

There are multiple ways to obtain the graph of g from the graph of f through a sequence of transformations. One choice is to use the structure of g(x) = -2 . 3^{x – 1}to translate the graph off horizontally one unit left, reflect across the horizontal axis, and then scale vertically by a factor of 2. Another choice is to rewrite g(x) as g(x) =Â \(\frac{2}{3}\)(3^{x}). Then the graph of g is obtained from the graph of f by reflecting the graph of f across the horizontal axis and then vertically scaling by a factor of \(\frac{2}{3}\).

### Eureka Math Algebra 2 Module 3 Lesson 20 Problem Set Answer Key

Question 1.

Describe each function as a transformation of the graph of a function in the form f(x) = log(x). Sketch the graph of f and the graph of g by hand. Label key features such as intercepts, intervals where g is increasing or decreasing, and the equation of the vertical asymptote.

a. g(x) = log_{2}(x – 3)

Answer:

The graph of g is the graph of f(x) = log_{2}(x) translated horizontally 3 units to the right. The function g is increasing on (3, âˆž). The x-intercept is 4, and the vertical asymptote is x = 3.

b. g(x) = log_{2}(16x)

Answer:

The graph of g is the graph of f(x) = log_{2}(x) translated vertically up 4 units. The function g is increasing on

(0, âˆž). The x-intercept is 2^{-4}. The vertical asymptote is x = 0.

c. g(x) = log_{2}(\(\frac{8}{x}\))

Answer:

The graph of g is the graph of f(x) = log_{2}(x) reflected about the horizontal axis and translated vertically up 3 units. The function g is decreasing on (0, âˆž). The x-intercept is 2^{3}. The vertical asymptote is x = 0. The reflected graph and the final graph are both shown in the solution.

d. g(x) = log_{2}((x – 3)^{2})for x > 3

Answer:

The graph of g is the graph of f(x) = log_{2}(x) stretched vertically by a factor of 2 and translated horizontally 3 units to the right. The function g is increasing on (3, âˆž). The x-intercept is 4, and the vertical asymptote is x = 3.

Question 2.

Each function graphed below can be expressed as a transformation of the graph of f(x) = log(x). Write an algebraic function for g and h, and state the domain and range.

Answer:

In Figure 1, g(x) = -log(x – 2) for x > 2. The domain of g is x > 2, and the range of g is all real numbers.

In Figure 2, h(x) = 2 + log(x) for x > 0. The domain of h is x > 0, and the range of h is all real numbers.

Figure 1: Graphs of f(x) = log(x) and the function g

Figure 2: Graphs of f(x) = log(x) and the function h

Question 3.

Describe each function as a transformation of the graph of a function in the form f(x) = b^{x}. Sketch the graph off and the graph of g by hand. Label key features such as intercepts, intervals where g is increasing or decreasing, and the horizontal asymptote.

a. g(x) = 2 . 3^{x} – 1

Answer:

The graph of g is the graph of f(x) = 3^{x} scaled vertically by a factor of 2 and translated vertically down 1 unit. The equation of the horizontal asymptote is y = -1. The y-intercept is 1, and the x-intercept is approximately -0.631. The function g is increasing for all real numbers.

b. g(x) = 2^{2x} + 3

Answer:

The graph of g is the graph of f(x) = 4^{x} translated vertically up 3 units. The equation of the horizontal asymptote is y = 3. The y-intercept is 4. There is no x-intercept. The function g is increasing for all real numbers.

c. g(x) = 3^{x – 2}

Answer:

The graph of g is the graph of f(x) = 3^{x} translated horizontally 2 units to the right OR the graph of f scaled

vertically by a factor of \(\frac{1}{9}\). The equation of the horizontal asymptote is y = 0. The y-intercept is \(\frac{1}{9}\). There is no x-intercept, and the function g is increasing for all real numbers.

d. g(x) = -9^{\(\frac{x}{2}\)} + 1

Answer:

The graph of g is the graph of f(x) = 3^{x} reflected about the horizontal axis and then translated vertically up 1 unit. The equation of the horizontal asymptote is y = 1. The y-intercept is 0, and the x-intercept is also 0. The function g is decreasing for all real numbers.

Question 4.

Using the function f(x) = 2^{x}, create a new function g whose graph is a series of transformations of the graph of f with the following characteristics:

â†’ The function g is decreasing for all real numbers.

â†’ The equation for the horizontal asymptote is y = 5.

â†’ The y-intercept is 7.

Answer:

One possible solution is g(x) = 2 . 2^{-x} + 5.

Question 5.

Using the function f(x) = 2^{x}, create a new function g whose graph is a series of transformations of the graph of f with the following characteristics:

â†’ The function g is increasing for all real numbers.

â†’ The equation for the horizontal asymptote is y = 5.

â†’ The y-intercept is 4.

Answer:

One possible solution is g(x) = -(2^{-x}) + 5.

Question 6.

Consider the function g(x) = (\(\frac{1}{4}\))^{x – 3}

a. Write the function g as an exponential function with base 4. Describe the transformations that would take the graph of f(x) = 4^{x} to the graph of g.

Answer:

(\(\frac{1}{4}\))^{x – 3} = (4^{-1})^{x – 3} = 4^{-x + 3} = 4^{3} . 4^{-x}

Thus, g(x) = 64 . 4^{-x}. The graph of g is the graph of f reflected about the vertical axis and scaled vertically

by a factor of 64.

b. Write the function g as an exponential function with base 2. Describe two different series of transformations that would take the graph of f(x) = 2â€™ to the graph of g.

Answer:

(\(\frac{1}{4}\))^{x – 3} = (2)^{-2}^{x – 3} = 2^{-2(x – 3)} = 2^{-2x + 6} = 64 . 2^{-2x}

Thus, g(x) = 64 . 2^{-2x} or g(x) = 2^{-2(x – 3)}. To obtain the graph of g from the graph of f, you can scale the graph horizontally by a factor of \(\frac{1}{2}\), reflect the graph about the vertical axis, and scale it vertically by a factor of 64. Or, you can scale the graph horizontally by a factor of \(\frac{1}{2}\), reflect the graph about the vertical axis, and translate the resulting graph horizontally 3 units to the right.

Question 7.

Explore the graphs of functions in the form f(x) = log(x^{n}) for n > 1. Explain how the graphs of these functions change as the values of n increase. Use a property of logarithms to support your reasoning.

Answer:

The graphs appear to be a vertical scaling of the common logarithm function by a factor of n. This is true because of the property of logarithms that states log(x^{n}) = n log(x).

Question 8.

Use a graphical approach to soâ€™ve each equation. If the equation has no solution, explain why.

a. log(x) = log(x – 2)

Answer:

This equation has no solution because the graphs of y = log(x) and y = log(x – 2) are horizontal translations of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution.

b. log(x) = log(2x)

Answer:

This equation has no solution because log(2x) = log(2) + log(x), which means that the graphs of y = log(x) andy = log(2x) are a vertical translation of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution.

c. log(x) = log (\(\frac{2}{x}\))

Ans:

The solution is the x-coordinate of the intersection point of the graphs of y = log(x) and y = log(2) – log(x). Since the graph of the function defined by the right side of the equation is a reflection across the horizontal axis and a vertical translation of the graph of the function defined by the left side of the equation, the graphs of these functions intersect in exactly one point with an x-coordinate approximately 1.5.

d. Show algebraically that the exact solution to the equation in part (c) is âˆš2.

Answer:

log(x) = log(2) – log(x)

2 log(x) = log(2)

log(x) = \(\frac{1}{2}\)log(2)

log(x) = log(2^{\(\frac{1}{2}\)})

x = 2^{\(\frac{1}{2}\)}

Since 2^{\(\frac{1}{2}\)} = âˆš2 the exact solution is âˆš2.

Question 9.

Make a table of values forf(x) = \(x^{\frac{1}{\log (x)}}\) for x > 1. Graph the function f for x > 1. Use properties of logarithms to explain what you see in the graph and the table of values.

Answer:

We see that \(x^{\frac{1}{\log (x)}}\) = 10 for all x > 1 because \(\frac{1}{\log (x)}=\frac{\log (10)}{\log (x)}\) = log_{x}(10). Therefore, when we substitute log_{x}(10) into the expression \(x^{\frac{1}{\log (x)}}\), we get x^{logx(10)}, which is equal to 10 according to the definition of a logarithm.

### Eureka Math Algebra 2 Module 3 Lesson 20 Exit Ticket Answer Key

Question 1.

Express g(x) = -log_{4}(2x) in the general form of a logarithmic function, f(x) = k + a log_{b}(x – h). Identify a, b, h, and k.

Answer:

Since -log_{4}(2x) = -log_{4}(2) + log_{4}(x) = –\(\frac{1}{2}\) – log_{4}(x), the function is g(x) = –\(\frac{1}{2}\) – log(x), and a = -1, b = 4, h = 0, and k = –\(\frac{1}{2}\).

Question 2.

Use the structure of g when written in general form to describe the graph of g as a transformation of the graph of h(x) = log_{4}(x).

Answer:

The graph of g is the graph of h reflected about the horizontal axis and translated down \(\frac{1}{2}\) unit.

Question 3.

Graph g and h on the same coordinate axes.

Answer:

The graph of h is shown in blue, and the graph of g is shown in green.