{"id":101969,"date":"2022-07-12T11:32:31","date_gmt":"2022-07-12T06:02:31","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=101969"},"modified":"2022-05-10T15:21:19","modified_gmt":"2022-05-10T09:51:19","slug":"eureka-math-precalculus-module-3-lesson-15","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-precalculus-module-3-lesson-15\/","title":{"rendered":"Eureka Math Precalculus Module 3 Lesson 15 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Precalculus Module 3 Lesson 15 Answer Key<\/h2>\n

Eureka Math Precalculus Module 3 Lesson 15 Example Answer Key<\/h3>\n

Example 1.
\nGraph the function f(x) = \\(\\frac{x + 2}{x – 3}\\) using transformations of the graph of y = \\(\\frac{1}{x}\\).
\nAnswer:
\nf(x) = \\(\\frac{x – 3 + 5}{x – 3}\\) = 1 + \\(\\frac{5}{x – 3}\\)
\n\"Engage<\/p>\n

Eureka Math Precalculus Module 3 Lesson 15 Exploratory Challenge\/Exercise Answer Key<\/h3>\n

Exploratory Challenge\/Exercises 1\u20132
\nExercise 1.
\nSketch the general shape of the graph of the function f(x) = \\(\\frac{1}{x^{n}}\\) for n > 0 when n is an odd number.
\nAnswer:
\n\"Engage<\/p>\n

Exercise 2.
\nSketch the general shape of the graph of the function f(x) = \\(\\frac{1}{x^{n}}\\) for n>0 when n is an even number.
\nAnswer:
\n\"Engage<\/p>\n

Exercises 3\u20135
\nExercise 3.
\nSketch 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.
\nAnswer:
\n\"Engage<\/p>\n

a. g(x) = \\(\\frac{1}{x – 2}\\)
\nAnswer:
\n\"Engage<\/p>\n

b. h(x) = – \\(\\frac{1}{x}\\) + 3
\nAnswer:
\n\"Engage<\/p>\n

c. k(x) = \\(\\frac{2}{x + 3}\\) – 5
\nAnswer:
\n\"Engage<\/p>\n

Exercise 4.
\nUse 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.
\nAnswer:
\na 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.<\/p>\n

Exercise 5.
\nSketch 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.
\nAnswer:
\n\"Engage<\/p>\n

a. g(x) = – \\(\\frac{3}{(x + 1)^{2}}\\)
\nAnswer:
\n\"Engage<\/p>\n

b. h(x) = \\(\\frac{1}{(x – 1)^{2}}\\) + 4
\nAnswer:
\n\"Engage<\/p>\n

Exercises 6\u201313
\nSketch 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.
\nExercise 6.
\nf(x) = \\(\\frac{x – 7}{x – 5}\\)
\nAnswer:
\nf(x) = 1 – \\(\\frac{2}{x – 5}\\)
\nThe graph is shifted right 5 units.
\n\"Engage<\/p>\n

Exercise 7.
\nf(x) = \\(\\frac{2 x + 6}{x + 1}\\)
\nAnswer:
\nf(x) = \\(\\frac{x + 1 + x + 1 + 4}{x + 1}\\) = 1 + 1 + \\(\\frac{4}{x + 1}\\) = 2 + \\(\\frac{4}{x + 1}\\)
\nThe graph is shifted left 1 unit and up 2 units.
\n\"Engage<\/p>\n

Exercise 8.
\nf(x) = \\(\\frac{2 x^{2} – 1}{x^{2}}\\)
\nAnswer:
\nf(x) = 2 – \\(\\frac{1}{x^{2}}\\)
\nThe graph is shifted up 2 units, and both branches are going down.
\n\"Engage<\/p>\n

Exercise 9.
\nf(x) = \\(\\frac{1 + 4 x^{3}}{x^{3}}\\)
\nAnswer:
\nf(x) = \\(\\frac{1}{x^{3}}\\) + 4
\nThe graph is shifted up 4 units.
\n\"Engage<\/p>\n

Exercise 10.
\nf(x) = \\(\\frac{x^{2} – 2 x + 3}{(x – 1)^{2}}\\)
\nAnswer:
\n\"Engage
\nThe graph is shifted right 1 unit and up 1 unit. Both branches go up.
\n\"Engage<\/p>\n

Exercise 11.
\nf(x) = \\(\\frac{2 x^{2} + 12 x + 13}{(x + 3)^{2}}\\)
\nAnswer:
\n\"Engage
\nThe graph is shifted left 3 units and up 2 units. Both branches go down.
\n\"Engage<\/p>\n

Exercise 12.
\nf(x) = \\(\\frac{x + 4}{x^{2} – 16}\\)
\nAnswer:
\nf(x) = \\(\\frac{1}{x – 4}\\), x\u2260 – 4
\nThe graph is shifted right 4 units, and there is a value missing at x = – 4.
\n\"Engage<\/p>\n

Exercise 13.
\nf(x) = \\(\\frac{x}{x^{3} – 4 x^{2} + 4 x}\\)
\nAnswer:
\nf(x) = \\(\\frac{1}{(x – 2)^{2}}\\), x\u22600
\nThe graph is shifted right 2 units, and there is a value missing at x = 0.
\nBoth branches go up.
\n\"Engage<\/p>\n

Eureka Math Precalculus Module 3 Lesson 15 Problem Set Answer Key<\/h3>\n

Question 1.
\nWrite 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}}\\).
\na. y = \\(\\frac{5 x – 8}{x + 2}\\)
\nAnswer:
\ny = \\(\\frac{5 x + 10 – 18}{x + 2}\\) = 5 – \\(\\frac{18}{x + 2}\\)
\nThe 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.<\/p>\n

b. y = \\(\\frac{2 x^{3} – 4}{x^{3}}\\)
\nAnswer:
\ny = 2 – \\(\\frac{4}{x^{3}}\\)
\nThe 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.<\/p>\n

c. y = \\(\\frac{x^{2} – 4 x + 8}{(x – 2)^{2}}\\)
\nAnswer:
\n\"Eureka
\nThe 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.<\/p>\n

d. y = \\(\\frac{3 x – 12}{x^{2} – 16}\\)
\nAnswer:
\n\"Eureka
\nThe 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.<\/p>\n

e. y = \\(\\frac{2 x^{2} + 16 x + 25}{x^{2} + 8 x + 16}\\)
\nAnswer:
\n\"Eureka
\nThe 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.<\/p>\n

Question 2.
\nFor each function in Problem 1, state how the horizontal and vertical asymptotes are affected from the original graph of y = \\(\\frac{1}{x^{n}}\\).
\nAnswer:
\na. The horizontal asymptote is moved up 5; the vertical asymptote is moved 2 to the left.
\nb. The horizontal asymptote is moved up 2; the vertical asymptote is unchanged.
\nc. The horizontal asymptote is moved up 1; the vertical asymptote is moved to the right 2.
\nd. The horizontal asymptote is unchanged; the vertical asymptote is moved to the left 4.
\ne. The horizontal asymptote is moved up 2; the vertical asymptote is moved to the left.<\/p>\n

Question 3.
\nSketch a picture of the graph of each function in Problem 1.
\nAnswer:
\na.
\n\"Eureka
\nb.
\n\"Eureka
\nc.
\n\"Eureka
\nd.
\n\"Eureka
\ne.
\n\"Eureka<\/p>\n

Question 4.
\nWhat are some indicators whether or not a rational function can be expressed as a transformation of y = \\(\\frac{1}{x^{n}}\\)?
\nAnswer:
\nIf 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.<\/p>\n

Question 5.
\nWrite 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.
\nAnswer:
\ny = \\(\\frac{2}{x – 2}\\) – 3<\/p>\n

Eureka Math Precalculus Module 3 Lesson 15 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nSketch 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.
\n\"Eureka
\nAnswer:
\ny = \\(\\frac{3 x – 7}{x – 3}\\)
\ny = \\(\\frac{3 x – 9 + 2}{x – 3}\\) = 3 + \\(\\frac{2}{x – 3}\\)
\nBy 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.
\n\"Eureka<\/p>\n","protected":false},"excerpt":{"rendered":"

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) = using transformations of the graph of y = . Answer: f(x) = = 1 + Eureka Math Precalculus Module 3 Lesson 15 Exploratory Challenge\/Exercise Answer Key Exploratory … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/101969"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=101969"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/101969\/revisions"}],"predecessor-version":[{"id":136193,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/101969\/revisions\/136193"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=101969"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=101969"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=101969"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}