2-5 Rational Functions

Transcription

1 -5 Rational Functions Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any 1 f () = The function is undefined at the real zeros of the denominator b() = 4 The real zeros of b() are and Therefore, D = {,, R} Check for vertical asymptotes Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f h() = The function is undefined at the real zero of the denominator b() = + 4 The real zero of b() is 4 Therefore, D = { 4, R} Check for vertical asymptotes Determine whether = 4 is a point of infinite discontinuity Find the limit as approaches 4 from the left and the right f () ,048 undef 71, Because = 4 is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, there does not appear to be a horizontal asymptote f () = The function is undefined at the real zeros of the denominator b() = ( + )( 4) The real zeros of b() are 4 and Therefore, D = { 4,, R} Page 1 Check for vertical asymptotes Determine whether = 4 is a point of infinite discontinuity Find the limit as approaches 4 from the left and the right

2 -5 Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () Rational Functions The table suggests Therefore, there does not appear to be a horizontal asymptote f () = The function is undefined at the real zeros of the denominator b() = ( + )( 4) The real zeros of b() are 4 and Therefore, D = { 4,, R} Check for vertical asymptotes Determine whether = 4 is a point of infinite discontinuity Find the limit as approaches 4 from the left and the right f () undef Because = 4 is a vertical asymptote of f Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, there does not appear to be a horizontal asymptote 4 g() = The function is undefined at the real zeros of the denominator b() = ( + )( + 5) The real zeros of b() are and 5 Therefore, D = {, 5, R} Check for vertical asymptotes Determine whether = 5 is a point of infinite discontinuity Find the limit as approaches 5 from the left and the right f () undef Because = 5 is a vertical asymptote of f Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f 5 h() = Page

3 -5 Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () Rational Functions The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f 5 h() = The function is undefined at the real zeros of the denominator b() = + The real zeros of b() are 0 and Therefore, D = { 0,, R} Check for vertical asymptotes Determine whether = 0 is a point of infinite discontinuity Find the limit as approaches 5 from the left and the right f () undef Because = 0 is a vertical asymptote of f Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () undef The table suggests Therefore, you know that y = is a horizontal asymptote of f 6 f () = The function is undefined at the real zero of the denominator b() = 4 The real zero of b() is 4 Therefore, D = { 4, R} Check for vertical asymptotes Determine whether = 4 is a point of infinite discontinuity Find the limit as approaches 4 from the left and the right f () ,98 undef 7, Because = 4 is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, there does not appear to be a horizontal asymptote 7 h() = The function is undefined at the real zeros of the denominator b() = ( )( + 4) The real zeros of b() are and 4 Therefore, D = {, 4, R} Check for vertical asymptotes Determine Cognero = is a point of infinite discontinuity Find the limit as approaches from the left and thepage esolutions Manual - whether Powered by right

4 -5 Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () Rational Functions The table suggests Therefore, there does not appear to be a horizontal asymptote 7 h() = The function is undefined at the real zeros of the denominator b() = ( )( + 4) The real zeros of b() are and 4 Therefore, D = {, 4, R} Check for vertical asymptotes Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Determine whether = 4 is a point of infinite discontinuity Find the limit as approaches 4 from the left and the right f () ,750 undef 416, Because = 4 is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f 8 g() = The function is undefined at the real zeros of the denominator b() = ( + 1)( ) The real zeros of b() are 1 and Therefore, D = {, 1, R} Check for vertical asymptotes Determine whether = is a point of infinite discontinuity Find the limit as approaches from the left and the right f () undef Because = is a vertical asymptote of f Determine whether = 1 is a point of infinite discontinuity Find the limit as approaches 1 from the left and the right f () undef Because = 1 is a vertical asymptote of f Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f For each function, determine any asymptotes and intercepts Then graph the function and state its domain 9 f () = Page 4

5 -5 Check for horizontal asymptotes Use a table to eamine the end behavior of f () f () Rational Functions The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f For each function, determine any asymptotes and intercepts Then graph the function and state its domain 9 f () = The function is undefined at b() = 0, so D = { 4, 5, R} There are vertical asymptotes at = 4 and = 5 There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercepts are and, the zeros of the numerator The y-intercept is because f (0) = Graph the asymptotes and intercepts Then find and plot points 10 g() = The function is undefined at b() = 0, so D = { 1,, R} There are vertical asymptotes at = and = 1 There is a horizontal asymptote at y =, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercepts are and 6, the zeros of the numerator The y-intercept is 9 because g(0) =9 Graph the asymptotes and intercepts Then find and plot points 11 f () = The function is undefined at b() = 0, so D = {,, R} Page 5 at = and = There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator esolutions Manual - Poweredasymptotes by Cognero There are vertical

6 -5 Rational Functions 11 f () = The function is undefined at b() = 0, so D = {,, R} There are vertical asymptotes at = and = There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator Since the polynomial in the numerator has no real zeros, there are no -intercepts The y-intercept is because f (0) = Graph the asymptotes and intercept Then find and plot points 1 f () = The function is undefined at b() = 0, so D = { 0, 6, R} There are vertical asymptotes at = 0 and = 6 There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is, the zero of the numerator There is no y-intercept because f () is undefined for = 0 Graph the asymptotes and intercept Then find and plot points 1 g() = g() can be simplified to g() = The function is undefined at b() = 0, so D = { 5, 6, R} There are vertical asymptotes at = 5 and = 6, the real zeros of the simplified denominator esolutions Manual Powered by asymptote Cognero There is a -horizontal at y = 0, because the degree of the denominator is greater than the degree of thepage 6 numerator The -intercept is, the zero of the simplified numerator The y-intercept is because

7 -5 Rational Functions 1 g() = g() can be simplified to g() = The function is undefined at b() = 0, so D = { 5, 6, R} There are vertical asymptotes at = 5 and = 6, the real zeros of the simplified denominator There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is, the zero of the simplified numerator The y-intercept is because Graph the asymptotes and intercepts Then find and plot points 14 h() = The function is undefined at b() = 0, so D = {, 0, 5, R} There are vertical asymptotes at =, = 0 and = 5 There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercepts are 6 and 4, the zeros of the numerator There is no y-intercept because h() is undefined for = 0 Graph the asymptotes and intercepts Then find and plot points 15 h() = esolutions h()manual can be- Powered writtenbyascognero h() = Page 7 The function is undefined at b() = 0, so D = {, 1, R}

8 -5 Rational Functions 15 h() = h() can be written as h() = The function is undefined at b() = 0, so D = {, 1, R} There are vertical asymptotes at = and = 1 There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator Since the degree of the numerator is greater than the denominator, there is no oblique asymptote The -intercepts are 5, 0, and, the zeros of the numerator The y-intercept is 0 because h(0) = 0 Graph the asymptotes and intercepts Then find and plot points 16 f () = f() can be written as f () = The function is undefined at b() = 0, so D = {, 8, R} There are vertical asymptotes at = 8 and = There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator Since the degree of the numerator is greater than the denominator, there is no oblique asymptote The -intercepts are 6, 0, and 4, the zeros of the numerator The y-intercept is 0 because f (0) = 0 Graph the asymptotes and intercepts Then find and plot points 17 f () = The function is undefined at b() = yields no real zeros, so D = { R} There are no vertical asymptotes esolutions Manual - Powered by Cognero Page 8 There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator

9 -5 Rational Functions 17 f () = The function is undefined at b() = yields no real zeros, so D = { R} There are no vertical asymptotes There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is 8, the zero of the numerator The y-intercept is because Graph the asymptote and intercepts Then find and plot points 18 The function is undefined at b() = yields no real zeros, so D = { R} There are no vertical asymptotes There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator Since the polynomial in the numerator has no real zeros, there are no -intercepts The y-intercept is because Graph the asymptote and intercept Then find and plot points 19 SALES The business plan for a new car wash projects that profits in thousands of dollars will be modeled by the function p (z) =, where z is the week of operation and z = 0 represents opening a State the domain of the function b Determine any vertical and horizontal asymptotes and intercepts for p (z) c Graph function esolutions Manualthe - Powered by Cognero Page 9

10 -5 Rational Functions 19 SALES The business plan for a new car wash projects that profits in thousands of dollars will be modeled by the function p (z) =, where z is the week of operation and z = 0 represents opening a State the domain of the function b Determine any vertical and horizontal asymptotes and intercepts for p (z) c Graph the function a The function is undefined at b() = 0 z + 7z + 5 can be written as (z + 5)(z+ 1) The zeros of this polynomial are at 1 and Since the car wash cannot be open for negative weeks, these zeros do not fall in the domain and D = {z z 0, z R} b Though there are vertical asymptotes at = 1 and, these values of are not in the domain Therefore, there are no vertical asymptotes for this situation There is a horizontal asymptote at y =, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercept is 1, the zero of the numerator The y-intercept is because c Find and plot points to construct a graph Since the domain is restricted to real numbers greater than or equal to 0, it is only necessary to show the first and fourth quadrants For each function, determine any asymptotes, holes, and intercepts Then graph the function and state its domain 0 h() = The function is undefined at b() = 0, so D = { 0, R} There is a vertical asymptote at = 0 There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is, the zero of the numerator There is no y-intercept because h() is undefined for = 0 Graph the asymptotes and intercept Then find and plot points Page 10

11 -5 Rational Functions For each function, determine any asymptotes, holes, and intercepts Then graph the function and state its domain 0 h() = The function is undefined at b() = 0, so D = { 0, R} There is a vertical asymptote at = 0 There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is, the zero of the numerator There is no y-intercept because h() is undefined for = 0 Graph the asymptotes and intercept Then find and plot points 1 h() = The function is undefined at b() = 0 Thus, There is a vertical asymptote at There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator Since the polynomial in the numerator has no real zeros, there are no -intercepts The y-intercept is (0) = because h Graph the asymptotes and intercept Then find and plot points Page 11

12 -5 Rational Functions 1 h() = The function is undefined at b() = 0 Thus, There is a vertical asymptote at There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator Since the polynomial in the numerator has no real zeros, there are no -intercepts The y-intercept is (0) = because h Graph the asymptotes and intercept Then find and plot points f () = f() can be written as f () = The function is undefined at b() = 0, so D = {, 1, R} There are vertical asymptotes at = and = 1 There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercepts are 5 and, the zeros of the numerator The y-intercept is 5 because f (0) = 5 Graph the asymptotes and intercepts Then find and plot points Page 1

13 -5 Rational Functions f () = f() can be written as f () = The function is undefined at b() = 0, so D = {, 1, R} There are vertical asymptotes at = and = 1 There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercepts are 5 and, the zeros of the numerator The y-intercept is 5 because f (0) = 5 Graph the asymptotes and intercepts Then find and plot points g() = The function is undefined at b() = 0, so D = { 4, R} There is a vertical asymptote at = 4 There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercept is 7, the zero of the numerator The y-intercept is because Graph the asymptotes and intercepts Then find and plot points 4 h() = The function is undefined at b() = 0, so D = {, R} There is a vertical asymptote at = There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator Page 1 Since the degree of the numerator is greater than the denominator, there is no oblique asymptote The -intercept is 0, the zero of the numerator The y-intercept is 0 because h(0) = 0 Graph the asymptote and intercepts Then find and plot points

14 -5 Rational Functions 4 h() = The function is undefined at b() = 0, so D = {, R} There is a vertical asymptote at = There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator Since the degree of the numerator is greater than the denominator, there is no oblique asymptote The -intercept is 0, the zero of the numerator The y-intercept is 0 because h(0) = 0 Graph the asymptote and intercepts Then find and plot points 5 g() = g() can be written as g() = The function is undefined at b() = 0, so D = { 4, R} There is a vertical asymptote at = 4 There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator Since the degree of the numerator is greater than the denominator, there is no oblique asymptote The -intercepts are, 1, and 0, the zeros of the numerator The y-intercept is 0 because g(0) = 0 Graph the asymptote and intercepts Then find and plot points 6 f () = f() can be written as f () = or The function is undefined at b() = 0, so D = {, 1,, R} There are vertical asymptotes at = 1 and =, the real zeros of the simplified denominator Page 14 There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator

15 -5 Rational Functions 6 f () = f() can be written as f () = or The function is undefined at b() = 0, so D = {, 1,, R} There are vertical asymptotes at = 1 and =, the real zeros of the simplified denominator There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is 7, the zero of the simplified numerator The y-intercept is because h(0) = There is a hole at (, 1) because the original function is undefined when = Graph the asymptotes and intercepts Then find and plot points 7 g() = or g() can be written as g() = The function is undefined at b() = 0, so D = {, 1, or, R} There is a vertical asymptote at = 1, the real zero of the simplified denominator There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator Since the simplified numerator has no real zeros, there are no -intercepts The y-intercept is 1 because g(0) =1 There are holes at (, 1) and because the original function is undefined when = and = Graph the asymptotes and intercept Then find and plot points 8 f () = esolutions Manual - Powered by Cognero f() can be written as f () = Page 15

16 -5 Rational Functions 8 f () = f() can be written as f () = The function is undefined at b() = 0, so D = {, 1, R} There is a vertical asymptote at =, the real zero of the simplified denominator There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal The -intercept is 4, the zero of the simplified numerator The y-intercept is There is a hole at because f (0) = because the original function is undefined when = 1 Graph the asymptotes and intercepts Then find and plot points 9 g() = g() can be written as g() = The function is undefined at b() = 0, so D = { 4, 5, R} There is a vertical asymptote at = 4, the real zero of the simplified denominator There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is, the zero of the simplified numerator The y-intercept is because g(0) = There is a hole at because the original function is undefined when = 5 Graph the asymptotes and intercepts Then find and plot points Page 16

17 -5 Rational Functions 9 g() = g() can be written as g() = The function is undefined at b() = 0, so D = { 4, 5, R} There is a vertical asymptote at = 4, the real zero of the simplified denominator There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator The -intercept is, the zero of the simplified numerator The y-intercept is because g(0) = There is a hole at because the original function is undefined when = 5 Graph the asymptotes and intercepts Then find and plot points 0 STATISTICS A number is said to be the harmonic mean of y and z if is the average of and a Write an equation for which the solution is the harmonic mean of 0 and 45 b Find the harmonic mean of 0 and 45 a Write an equation that represents as the average of and = Substitute y = 0 and z = 45 b Solve for for the equation found in part a Page 17

18 -5 Rational Functions 0 STATISTICS A number is said to be the harmonic mean of y and z if is the average of and a Write an equation for which the solution is the harmonic mean of 0 and 45 b Find the harmonic mean of 0 and 45 a Write an equation that represents as the average of and = Substitute y = 0 and z = 45 b Solve for for the equation found in part a The harmonic mean of 0 and 45 is 6 1 OPTICS The lens equation is = +, where f is the focal length, d i is the distance from the lens to the image, and d o is the distance from the lens to the object Suppose the object is centimeters from the lens and the focal length is 8 centimeters a Write a rational equation to model the situation b Find the distance from the lens to the image a Substitute f = 8 and d o = into the equation esolutions Manualfor - Powered by Cognero b Solve d for the equation i found in part a Page 18

19 -5 Rational Functions The harmonic mean of 0 and 45 is 6 1 OPTICS The lens equation is = +, where f is the focal length, d i is the distance from the lens to the image, and d o is the distance from the lens to the object Suppose the object is centimeters from the lens and the focal length is 8 centimeters a Write a rational equation to model the situation b Find the distance from the lens to the image a Substitute f = 8 and d o = into the equation b Solve for d i for the equation found in part a The distance from the lens to the image is 10 centimeters Solve each equation y + = 5 y = or y = z = 4 Page 19

20 -5 yrational = or y = Functions z = = 1 = 1 or = 8 5 Page 0

21 -5 Rational Functions = 1 or = 8 5 Because the original equation is not defined when y =, you can eliminate this etraneous solution So, has no solution 6 + = 7 Page 1

22 -5 Rational Functions 7 8 = 4 or = 5 9 Page

23 -5 Rational Functions = 4 or = = = 7 or = 1 41 Page

24 -5 Rational Functions = 7 or = WATER The cost per day to remove percent of the salt from seawater at a desalination plant is c() =, where 0 < 100 a Graph the function using a graphing calculator b Graph the line y = 8000 and find the intersection with the graph of c() to determine what percent of salt can be removed for $8000 per day c According to the model, is it feasible for the plant to remove 100% of the salt? Eplain your reasoning a b Use the CALC menu to find the intersection of c() and the line y = 8000 For $8000 per day, about 889% of salt can be removed c No; sample answer: The function is not defined when = 100 This suggests that it is not financially feasible to remove 100% of the salt at the plant Write a rational function for each set of characteristics 4 -intercepts at = 0 and = 4, vertical asymptotes at = 1 and = 6, and a horizontal asymptote at y = 0 Page 4

25 -5 Rational Functions 4 WATER The cost per day to remove percent of the salt from seawater at a desalination plant is c() =, where 0 < 100 a Graph the function using a graphing calculator b Graph the line y = 8000 and find the intersection with the graph of c() to determine what percent of salt can be removed for $8000 per day c According to the model, is it feasible for the plant to remove 100% of the salt? Eplain your reasoning a b Use the CALC menu to find the intersection of c() and the line y = 8000 For $8000 per day, about 889% of salt can be removed c No; sample answer: The function is not defined when = 100 This suggests that it is not financially feasible to remove 100% of the salt at the plant Write a rational function for each set of characteristics 4 -intercepts at = 0 and = 4, vertical asymptotes at = 1 and = 6, and a horizontal asymptote at y = 0 Sample answer: For the function to have -intercepts at = 0 and = 4, the zeros of the numerator need to be 0 and 4Thus, factors of the numerator are and ( 4) For there to be vertical asymptotes at = 1 and = 6, the zeros of the denominator need to be 1 and 6 Thus, factors of the denominator are ( 1) and ( 6) For there to be a horizontal asymptote at y = 0, the degree of the denominator needs to be greater than the degree of the numerator Raising the factor ( 6) to the second power will result in an asymptote at y = 0 f() = 44 -intercepts at = and =, vertical asymptote at = 4, and point discontinuity at ( 5, 0) Sample answer: For the function to have -intercepts at = and =, the zeros of the numerator need to be and Thus, factors of the numerator are ( ) and ( + ) For there to be a vertical asymptote at = 4, a zero of the denominator needs to be 4 Thus, a factor of the denominator is ( 4) For there to be point discontinuity at = 5, ( + 5) must be a factor of both the numerator and denominator Additionally, for the point discontinuity topage be 5 at ( 5, 0), the numerator must have a zero at = 5 Thus, the ( + 5) in the numerator must have a power of so that = 5 remains a zero of the numerator after the function is simplified

26 horizontal asymptote at y = 0, the degree of the denominator needs to be greater than the degree of the numerator Raising the factor ( 6) to the second power will result in an asymptote at y = 0 f() = -5 Rational Functions 44 -intercepts at = and =, vertical asymptote at = 4, and point discontinuity at ( 5, 0) Sample answer: For the function to have -intercepts at = and =, the zeros of the numerator need to be and Thus, factors of the numerator are ( ) and ( + ) For there to be a vertical asymptote at = 4, a zero of the denominator needs to be 4 Thus, a factor of the denominator is ( 4) For there to be point discontinuity at = 5, ( + 5) must be a factor of both the numerator and denominator Additionally, for the point discontinuity to be at ( 5, 0), the numerator must have a zero at = 5 Thus, the ( + 5) in the numerator must have a power of so that = 5 remains a zero of the numerator after the function is simplified f() = 45 TRAVEL When distance and time are held constant, the average rates, in miles per hour, during a round trip can be modeled by, where r1 represents the average rate during the first leg of the trip and r represents the average rate during the return trip a Find the vertical and horizontal asymptotes of the function, if any Verify your answer graphically b Copy and complete the table shown c Is a domain of r1 > 0 reasonable for this situation? Eplain your reasoning a There is a vertical asymptote at the real zero of the denominator r1 = 0 There is a horizontal asymptote at r = or r = 0, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal b Evaluate the function for each value of r1 to determine r Page 6

27 at ( 5, 0), the numerator must have a zero at = 5 Thus, the ( + 5) in the numerator must have a power of so that = 5 remains a zero of the numerator after the function is simplified = -5 f() Rational Functions 45 TRAVEL When distance and time are held constant, the average rates, in miles per hour, during a round trip can be modeled by, where r1 represents the average rate during the first leg of the trip and r represents the average rate during the return trip a Find the vertical and horizontal asymptotes of the function, if any Verify your answer graphically b Copy and complete the table shown c Is a domain of r1 > 0 reasonable for this situation? Eplain your reasoning a There is a vertical asymptote at the real zero of the denominator r1 = 0 There is a horizontal asymptote at r = or r = 0, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal b Evaluate the function for each value of r1 to determine r c No; sample answer: As r1 approaches infinity, r approaches 0 This suggests that the average speeds reached during the first leg of the trip have no bounds Being able to reach an infinite speed is not reasonable The same holds true about r as r1 approaches 0 from the right Use your knowledge of asymptotes and the provided points to epress the function represented by each graph Page 7

28 c No; sample answer: As r1 approaches infinity, r approaches 0 This suggests that the average speeds reached during the first leg of the trip have no bounds Being able to reach an infinite speed is not reasonable The same -5 holds Rational Functions true about r as r approaches 0 from the right 1 Use your knowledge of asymptotes and the provided points to epress the function represented by each graph 46 Sample answer: Since there are vertical asymptotes at = 6 and = 1, ( + 6) and ( 1) are factors of the denominator Since = 4 is a -intercept, ( 4) is a factor of the numerator Because there is a horizontal asymptote at y = 1, the degrees of the polynomials in the numerator and denominator are equal and the ratio of their leading coefficients is 1 There also appears to be an additional -intercept that is not labeled Let = a be the second -intercept ( a) is then a factor of the numerator A function that meets these characteristics is f() = Substitute the point ( 5, 6) into the equation and solve for a Substitute a = 1 into the original equation A function that can represent the graph is f () = 47 Page 8

29 function that can represent the graph is f () = -5 A Rational Functions 47 Sample answer: Since there are vertical asymptotes at = 4 and =, ( + 4) and ( ) are factors of the denominator Since = 5 is a -intercept, ( 5) is a factor of the numerator Because there is a horizontal asymptote at y =, the degrees of the polynomials in the numerator and denominator are equal and the ratio of their leading coefficients is There also appears to be an additional -intercept that is not labeled Let = a be the second -intercept ( a) is then a factor of the numerator A function that meets these characteristics is f() = Substitute the point ( 1, ) into the equation and solve for a Substitute a = into the original equation A function that can represent the graph is f () = Use the intersection feature of a graphing calculator to solve each equation 48 =8 Graph y = graphs and y = 8 using a graphing calculator Use the CALC menu to find the intersection ofpage the 9

30 -5 A Rational Functions function that can represent the graph is f () = Use the intersection feature of a graphing calculator to solve each equation 48 =8 and y = 8 using a graphing calculator Use the CALC menu to find the intersection of the Graph y = graphs The intersections occur at 159 and 1005 Thus, the solutions to = 8 are about 159 and about =1 and y = 1 using a graphing calculator Use the CALC menu to find the intersection of the Graph y = graphs The intersections occur at 65 and 56 Thus, the solutions to = 1 are about 65 and about = Graph y = and y = using a graphing calculator Use the CALC menu to find the intersection of the graphs Page 0

31 The intersections occur at 65 and 56 Thus, the solutions to -5 Rational Functions = 1 are about 65 and about = and y = using a graphing calculator Use the CALC menu to find the intersection of the Graph y = graphs The intersections occur at 098 and 090 Thus, the solutions to = are about 098 and about =6 Graph y = and y = 6 using a graphing calculator Use the CALC menu to find the intersection of the graphs The intersections occur at 87, 10, and 70 Thus, the solutions to = 6 are about 87, about 10, and about 70 5 CHEMISTRY When a 60% acetic acid solution is added to 10 liters of a 0% acetic acid solution in a 100-liter tank, the concentration of the total solution changes a Show that the concentration of the solution is f (a) =, where a is the volume of the 60% solution b Find the relevant domain of f (a) and the vertical or horizontal asymptotes, if any Page 1

32 The intersections occur at 87, 10, and 70 Thus, the solutions to = 6 are about -5 Rational Functions 87, about 10, and about 70 5 CHEMISTRY When a 60% acetic acid solution is added to 10 liters of a 0% acetic acid solution in a 100-liter tank, the concentration of the total solution changes a Show that the concentration of the solution is f (a) =, where a is the volume of the 60% solution b Find the relevant domain of f (a) and the vertical or horizontal asymptotes, if any c Eplain the significance of any domain restrictions or asymptotes d Disregarding domain restrictions, are there any additional asymptotes of the function? Eplain a Sample answer: The concentration of the total solution is the sum of the amount of acetic acid in the original 10 liters and the amount in the a liters of the 60% solution, divided by the total amount of solution or Multiplying both the numerator and the denominator by 5 gives or b Since a cannot be negative and the maimum capacity of the tank is 100 liters, the relevant domain of a is 0 a 90 There is a horizontal asymptote at y = 06, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal c Sample answer: Because the tank already has 10 liters of solution in it and it will only hold a total of 100 liters, the amount of solution added must be less than or equal to 90 liters It is also impossible to add negative amounts of solution, so the amount added must be greater than or equal to 0 As you add more of the 60% solution, the concentration of the total solution will get closer to 60%, but because the solution already in the tank has a lower concentration, the concentration of the total solution can never reach 60% Therefore, there is a horizontal asymptote at y = 06 d Yes; sample answer: The function is not defined at a = 10, but because the value is not in the relevant domain, the asymptote does not pertain to the function If there were no domain restrictions, there would be a vertical asymptote at a = 10 5 MULTIPLE REPRESENTATIONS In this problem, you will investigate asymptotes of rational functions a TABULAR Copy and complete the table below Determine the horizontal asymptote of each function algebraically b GRAPHICAL Graph each function and its horizontal asymptote from part a c TABULAR Copy and complete the table below Use the Rational Zero Function to help you find the real zeros of the numerator of each function Page

33 GRAPHICAL Graph each function and its horizontal asymptote from part a -5 b Rational Functions c TABULAR Copy and complete the table below Use the Rational Zero Function to help you find the real zeros of the numerator of each function d VERBAL Make a conjecture about the behavior of the graph of a rational function when the degree of the denominator is greater than the degree of the numerator and the numerator has at least one real zero a Since the degree of the denominator is greater than the degree of the numerator, f (), h(), and g() will have horizontal asymptotes at y = 0 b Use a graphing calculator to graph f (), h(), and g() c c To find the real zeros of the numerator of f (), use the Rational Zero Theorem and factoring Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4 Therefore, the possible rational zeros of f are ±1, ±, and ±4 Factoring results in ( 4)( 1) Thus, the real zeros of the numerator of f () are 4 and 1 To find the real zeros of the numerator of h(), use the Rational Zero Theorem and synthetic division Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 1 Therefore, the possible rational zeros of f are ±1, ±, ±, ±4, ±6, and ±1 By using synthetic division, it can be determined that = is a rational zero TheManual remaing quadratic factor esolutions - Powered by Cognero ( + 4) yields no real zeros Thus, the real zero of the numerator of h() is Page To find the real zeros of the numerator of g(), use the Rational Zero Theorem and factoring Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 1 Therefore, the possible

34 possible rational zeros of f are ±1, ±, ±, ±4, ±6, and ±1 By using synthetic division, it can be determined that = is a rational zero -5 Rational Functions The remaing quadratic factor ( + 4) yields no real zeros Thus, the real zero of the numerator of h() is To find the real zeros of the numerator of g(), use the Rational Zero Theorem and factoring Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 1 Therefore, the possible 4 rational zeros of f are ±1 Factoring 1 results in ( 1)( + 1) or ( + 1)( 1)( + 1) The quadratic factor ( + 1) yields no real zeros Thus, the real zeros of the numerator of g() are 1 and 1 d Sample answer: When the degree of the numerator is less than the degree of the denominator and the numerator has at least on real zero, the graph of the function will have y = 0 as an asymptote and will intersect the asymptote at the real zeros of the numerator 54 REASONING Given f () =, will f () sometimes, always, or never have a horizontal asymptote at y = 1 if a, b, c, d, e, and f are constants with a 0 and d 0? Eplain Sometimes; sample answer: When a = d, the function will have a horizontal asymptote at y = 1 When a d, the function will not have a horizontal asymptote at y = 1 55 PREWRITE Design a lesson plan to teach the graphing rational functions topics in this lesson Make a plan that addresses purpose, audience, a controlling idea, logical sequence, and time frame for completion See students work 56 CHALLENGE Write a rational function that has vertical asymptotes at = and = and an oblique asymptote y = Sample answer: Since there are vertical asymptotes at = and =, ( + ) and ( ) are factors of the denominator Since there is an oblique asymptote y =, the degree of the numerator is eactly 1 greater than the degree of the denominator Additionally, when the numerator is divided by the denominator, the quotient polynomial q () is q() = So, the function can be written as f () = = + function We know that, where a() is the numerator of the, where r() is the remainder We can write the denominator of f () as 6 and can use the equation to solve for a() Page 4

35 55 PREWRITE Design a lesson plan to teach the graphing rational functions topics in this lesson Make a plan that addresses purpose, audience, a controlling idea, logical sequence, and time frame for completion -5 Rational Functions See students work 56 CHALLENGE Write a rational function that has vertical asymptotes at = and = and an oblique asymptote y = Sample answer: Since there are vertical asymptotes at = and =, ( + ) and ( ) are factors of the denominator Since there is an oblique asymptote y =, the degree of the numerator is eactly 1 greater than the degree of the denominator Additionally, when the numerator is divided by the denominator, the quotient polynomial q () is q() = So, the function can be written as f () = = + function We know that, where a() is the numerator of the, where r() is the remainder We can write the denominator of f () as 6 and can use the equation to solve for a() The degree of the remainder has to be less than the degree of the denominator, so it will either be 1 or 0 Thus, the sum of 18 + r() cannot be determined, but the first two terms of a() must be Substitute this epression for a() and use long division to verify that the quotient is The quotient is There is a remainder but it has no bearing on quotient, which is the oblique asymptote Thus, a function that has vertical asymptotes at = and = and an oblique asymptote y = is f () = 57 Writing in Math Use words, graphs, tables, and equations to show how to graph a rational function See students work 58 CHALLENGE Solve for k so that the rational equation has eactly one etraneous solution and one real solution = + Let k = so that the denominator factors to ( 1)( ) Page 5

36 57 Writing in Math Use words, graphs, tables, and equations to show how to graph a rational function -5 Rational Functions See students work 58 CHALLENGE Solve for k so that the rational equation has eactly one etraneous solution and one real solution = + Let k = so that the denominator factors to ( 1)( ) The solutions are = and Because the original equation is not defined when =, this is the etraneous solution Thus, when k =, the rational equation has eactly one etraneous solution and one real solution 59 Writing in Math Eplain why all of the test intervals must be used in order to get an accurate graph of a rational function Sample answer: The test intervals are used to determine the location of points on the graph Because many rational functions are not continuous, one interval may include y-values that are vastly different than the net interval Therefore, at least one, and preferably more than one, point is needed for every interval in order to sketch a reasonably accurate graph of the function List all the possible rational zeros of each function Then determine which, if any, are zeros 60 f () = Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 6 Therefore, the possible rational zeros of g are ±1, ±, ±, and ±6 By using synthetic division, it can be determined that = 1 is a rational zero Page 6 Because ( + 1) is a factor of f (), we can use the quotient to write a factored form of f () as f () = ( + 1)( +

37 -5 Sample answer: The test intervals are used to determine the location of points on the graph Because many rational functions are not continuous, one interval may include y-values that are vastly different than the net interval Therefore, least one, and preferably more than one, point is needed for every interval in order to sketch a RationalatFunctions reasonably accurate graph of the function List all the possible rational zeros of each function Then determine which, if any, are zeros 60 f () = Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 6 Therefore, the possible rational zeros of g are ±1, ±, ±, and ±6 By using synthetic division, it can be determined that = 1 is a rational zero Because ( + 1) is a factor of f (), we can use the quotient to write a factored form of f () as f () = ( + 1)( + 6) Factoring the quadratic epression yields f () = ( + 1)( + )( ) Thus, the rational zeros of f are 1,, and 61 f () = Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18 Therefore, the possible rational zeros of g are ±1, ±, ±, ±6, ±9, and ±18 By using synthetic division, it can be determined that = is a rational zero Because ( + ) is a factor of f (), we can use the quotient to write a factored form of f () as f () = ( + )( 4 + 9) The quadratic epression ( 4 + 9) yields no rational zeros Thus, the rational zero of f is 6 f () = Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term Therefore, the possible rational zeros of g are ±1 and ± By using synthetic division, it can be determined that = 1 is a rational zero By using synthetic division on the depressed polynomial, it can be determined that = is a rational zero Because ( 1) and ( ) are factors of f (), we can use the final quotient to write a factored form of f () as f () = ( 1)( )( + 1) Factoring the quadratic epression yields f () = ( 1)( )( 1)( 1) or f () = ( 1) ( ) Thus, the rational zeros of f are 1 (multiplicity: ) and Use the Factor Theorem to determine if the binomials given are factors of f () Use the binomials that are factors to write a factored form of f () 6 f () = ;, Use synthetic division to test each factor, ( ) and ( ) Page 7

38 Because ( 1) and ( ) are factors of f (), we can use the final quotient to write a factored form of f () as f () = ( 1)( )( + 1) Factoring the quadratic epression yields f () = ( 1)( )( 1)( 1) or f () = -5 Rational Functions ( 1) ( ) Thus, the rational zeros of f are 1 (multiplicity: ) and Use the Factor Theorem to determine if the binomials given are factors of f () Use the binomials that are factors to write a factored form of f () 6 f () = ;, Use synthetic division to test each factor, ( ) and ( ) Because the remainder when f () is divided by ( ) is 7, ( ) is not a factor Test the second factor, ( ) Because the remainder when f () is divided by ( ) is 0, ( ) is a factor Because ( ) is a factor of f (), we can use the quotient of f () ( ) to write a factored form of f () as f () = ( )( 1 1) 64 f () = ; 4, 1 Use synthetic division to test each factor, ( 4) and ( 1) Because the remainder when f () is divided by ( 4) is 0, ( 4) is a factor For ( 1), use the depressed polynomial + + and rewrite the division epression so that the divisor is of the form c Because Set up the synthetic division as follows Then follow the synthetic division procedure Because the remainder when the depressed polynomial is divided by ( 1) is, ( 1) is not a factor Because ( 4) is a factor of f (), we can use the quotient of f () ( 4) to write a factored form of f () as fpage () 8 = ( 4)( + + ) Factoring the cubic epression + + results in ( + + 1) or ( + 1)( + 1) Thus, f () = ( 4)( + + ) or f () = ( + 1)( + 1)( 4)

39 Because the remainder when f () is divided by ( ) is 0, ( ) is a factor Because ( ) is a factor of f (), we can use the quotient of f () ( ) to write a factored form of f () as f () -5 =Rational ( )( Functions 1 1) 64 f () = ; 4, 1 Use synthetic division to test each factor, ( 4) and ( 1) Because the remainder when f () is divided by ( 4) is 0, ( 4) is a factor For ( 1), use the depressed polynomial + + and rewrite the division epression so that the divisor is of the form c Because Set up the synthetic division as follows Then follow the synthetic division procedure Because the remainder when the depressed polynomial is divided by ( 1) is, ( 1) is not a factor Because ( 4) is a factor of f (), we can use the quotient of f () ( 4) to write a factored form of f () as f () = ( 4)( + + ) Factoring the cubic epression + + results in ( + + 1) or ( + 1)( + 1) Thus, f () = ( 4)( + + ) or f () = ( + 1)( + 1)( 4) 65 f () = ;, 5 Use synthetic division to test each factor, ( ) and ( 5) For ( ), rewrite the division epression so that the divisor is of the form c Because Set up the synthetic division as follows Then follow the synthetic division procedure Page 9

40 Because the remainder when the depressed polynomial is divided by ( 1) is, ( 1) is not a factor Because ( 4) is a factor of f (), we can use the quotient of f () ( 4) to write a factored form of f () as f () = ( 4)( + + ) Factoring the cubic epression + + results in ( + + 1) or ( + 1)( + 1) -5 Rational Functions Thus, f () = ( 4)( + + ) or f () = ( + 1)( + 1)( 4) 65 f () = ;, 5 Use synthetic division to test each factor, ( ) and ( 5) For ( ), rewrite the division epression so that the divisor is of the form c Because Set up the synthetic division as follows Then follow the synthetic division procedure Because the remainder when f () is divided by ( ) is 0, ( ) is a factor Test the second factor, ( 5), with the depressed polynomial Because the remainder when the depressed polynomial is divided by ( 5) is 1100, ( 5) is not a factor of f () Because ( ) is a factor of f (), we can use the quotient of f () ( ) to write a factored form of f () as f () = ( )( ) Synthetic division can be used to factor By using synthetic division, it can be determined that = 5 is a rational zero The remaining quadratic epression ( ) can be written as ( + 5)( + 1) Thus, f () = ( ) ( ) or f () = ( )( + 1)( + 5) 66 f () = ; 4 ; 1 Use synthetic division to test each factor, (4 ) and ( 1) For (4 ), rewrite the division epression so that the divisor is of the form c Because Set up the synthetic division as follows Then follow the synthetic division procedure Page 40