# Eureka Math Precalculus Module 3 Lesson 15 Answer Key

## Engage NY Eureka Math Precalculus Module 3 Lesson 15 Answer Key

### Eureka Math Precalculus Module 3 Lesson 15 Example Answer Key

Example 1.
Graph the function f(x) = $$\frac{x + 2}{x – 3}$$ using transformations of the graph of y = $$\frac{1}{x}$$.
f(x) = $$\frac{x – 3 + 5}{x – 3}$$ = 1 + $$\frac{5}{x – 3}$$ ### Eureka Math Precalculus Module 3 Lesson 15 Exploratory Challenge/Exercise Answer Key

Exploratory Challenge/Exercises 1–2
Exercise 1.
Sketch the general shape of the graph of the function f(x) = $$\frac{1}{x^{n}}$$ for n > 0 when n is an odd number. Exercise 2.
Sketch the general shape of the graph of the function f(x) = $$\frac{1}{x^{n}}$$ for n>0 when n is an even number. Exercises 3–5
Exercise 3.
Sketch the graph of the function f(x) = $$\frac{1}{x}$$. Then, use the graph of f to sketch each transformation of f showing the vertical and horizontal asymptotes. a. g(x) = $$\frac{1}{x – 2}$$ b. h(x) = – $$\frac{1}{x}$$ + 3 c. k(x) = $$\frac{2}{x + 3}$$ – 5 Exercise 4.
Use your results from Exercise 3 to make some general statements about graphs of functions in the form f(x) = a + $$\frac{b}{x – c}$$. Describe the effect that changing each parameter a,b,and c has on the graph of f.
a represents the vertical shift. A positive value of a shifts the graph up a units, and a negative value shifts the graph down a units. The horizontal asymptote is y = a. c represents the horizontal shift. A positive value of c shifts the graph right c units, and a negative value shifts the graph left c units. The vertical asymptote is x = c. b represents the vertical scaling. For values of b such that |b| > 1, the graph stretches vertically. For values of b such that 0 < |b| < 1, the graph compresses vertically. When b is negative, the graph reflects across the x – axis.

Exercise 5.
Sketch the graph of the function f(x) = $$\frac{1}{x^{2}}$$. Then, use the graph of f to sketch each transformation of f showing the vertical and horizontal asymptotes. a. g(x) = – $$\frac{3}{(x + 1)^{2}}$$ b. h(x) = $$\frac{1}{(x – 1)^{2}}$$ + 4 Exercises 6–13
Sketch each function by using transformations of the graph of y = $$\frac{1}{x}$$ or the graph of y = $$\frac{1}{x^{2}}$$. Explain the transformations that are evident in each example.
Exercise 6.
f(x) = $$\frac{x – 7}{x – 5}$$
f(x) = 1 – $$\frac{2}{x – 5}$$
The graph is shifted right 5 units. Exercise 7.
f(x) = $$\frac{2 x + 6}{x + 1}$$
f(x) = $$\frac{x + 1 + x + 1 + 4}{x + 1}$$ = 1 + 1 + $$\frac{4}{x + 1}$$ = 2 + $$\frac{4}{x + 1}$$
The graph is shifted left 1 unit and up 2 units. Exercise 8.
f(x) = $$\frac{2 x^{2} – 1}{x^{2}}$$
f(x) = 2 – $$\frac{1}{x^{2}}$$
The graph is shifted up 2 units, and both branches are going down. Exercise 9.
f(x) = $$\frac{1 + 4 x^{3}}{x^{3}}$$
f(x) = $$\frac{1}{x^{3}}$$ + 4
The graph is shifted up 4 units. Exercise 10.
f(x) = $$\frac{x^{2} – 2 x + 3}{(x – 1)^{2}}$$ The graph is shifted right 1 unit and up 1 unit. Both branches go up. Exercise 11.
f(x) = $$\frac{2 x^{2} + 12 x + 13}{(x + 3)^{2}}$$ The graph is shifted left 3 units and up 2 units. Both branches go down. Exercise 12.
f(x) = $$\frac{x + 4}{x^{2} – 16}$$
f(x) = $$\frac{1}{x – 4}$$, x≠ – 4
The graph is shifted right 4 units, and there is a value missing at x = – 4. Exercise 13.
f(x) = $$\frac{x}{x^{3} – 4 x^{2} + 4 x}$$
f(x) = $$\frac{1}{(x – 2)^{2}}$$, x≠0
The graph is shifted right 2 units, and there is a value missing at x = 0.
Both branches go up. ### Eureka Math Precalculus Module 3 Lesson 15 Problem Set Answer Key

Question 1.
Write each function so that it appears to be a transformation of y = $$\frac{1}{x^{n}}$$. Then, explain how the graph of each function relates to the graph of y = $$\frac{1}{x^{n}}$$.
a. y = $$\frac{5 x – 8}{x + 2}$$
y = $$\frac{5 x + 10 – 18}{x + 2}$$ = 5 – $$\frac{18}{x + 2}$$
The graph would be the graph of y = $$\frac{1}{x}$$ shifted left 2 units, stretched vertically by a scale factor of 18, reflected across the x – axis, and shifted up 5 units.

b. y = $$\frac{2 x^{3} – 4}{x^{3}}$$
y = 2 – $$\frac{4}{x^{3}}$$
The graph would be the graph of y = $$\frac{1}{x^{3}}$$ stretched vertically by a scale factor of 4, reflected across the x – axis, and shifted up 2 units.

c. y = $$\frac{x^{2} – 4 x + 8}{(x – 2)^{2}}$$ The graph would be the graph of y = $$\frac{1}{x^{2}}$$ shifted right 2 units, stretched vertically by a scale factor of 4, reflected across the x – axis, and shifted up 1 unit.

d. y = $$\frac{3 x – 12}{x^{2} – 16}$$ The graph would be the graph of y = $$\frac{1}{x}$$ shifted left 4 units and stretched vertically by a scale factor of 3. The point at x = 4 would be missing from the graph.

e. y = $$\frac{2 x^{2} + 16 x + 25}{x^{2} + 8 x + 16}$$ The graph would be the graph of y = $$\frac{1}{x^{2}}$$ shifted left 4 units, stretched vertically by a scale factor of 7, reflected across the x – axis, and shifted up 2 units.

Question 2.
For each function in Problem 1, state how the horizontal and vertical asymptotes are affected from the original graph of y = $$\frac{1}{x^{n}}$$.
a. The horizontal asymptote is moved up 5; the vertical asymptote is moved 2 to the left.
b. The horizontal asymptote is moved up 2; the vertical asymptote is unchanged.
c. The horizontal asymptote is moved up 1; the vertical asymptote is moved to the right 2.
d. The horizontal asymptote is unchanged; the vertical asymptote is moved to the left 4.
e. The horizontal asymptote is moved up 2; the vertical asymptote is moved to the left.

Question 3.
Sketch a picture of the graph of each function in Problem 1.
a. b. c. d. e. Question 4.
What are some indicators whether or not a rational function can be expressed as a transformation of y = $$\frac{1}{x^{n}}$$?
If the variables in the numerator can be expressed as a part of a multiple of the denominator, then the function is able to be expressed as a transformation. If it cannot, then there is a variable term in the numerator that cannot be removed that is the result of a composition or product of functions and not from a transformation.

Question 5.
Write an equation for a function whose graph is a transformation of the graph y = $$\frac{1}{x}$$. The graph has been shifted right 2 units, stretched vertically by a factor of 2, and shifted down 3 units.
y = $$\frac{2}{x – 2}$$ – 3

### Eureka Math Precalculus Module 3 Lesson 15 Exit Ticket Answer Key

Question 1.
Sketch the graph of the function given below by using transformations of y = $$\frac{1}{x^{n}}$$ . Explain which transformations you used and how you identified them. y = $$\frac{3 x – 7}{x – 3}$$
y = $$\frac{3 x – 9 + 2}{x – 3}$$ = 3 + $$\frac{2}{x – 3}$$
By rewriting the function into the form a + $$\frac{b}{x-c}$$, I saw that the graph was shifted right 3 units, stretched vertically by a factor of 2, and shifted up 3 units. 