Math 111 Lecture Notes Section 3.3: Graphing Rational Functions

Math 111 Lecture Notes Section 3.3: Graphing Rational Functions A rational function is of the form R() = p() q() where p and q are polnomial functions. The zeros of a rational function occur where p() = 0, as the function s value is zero where the value of the numerator is zero. Most often, the zeros will occur at a when the factor ( a) is in the numerator of R. A rational function is undefined where q() = 0, as the function is undefined whenever its denominator is zero. A vertical asmptote occurs when the denominator of the simplified form of R is equal to zero. Most often, the vertical asmptote = b will occur when the factor ( b) is in the denominator of the simplified form of R. A hole occurs when both the numerator and denominator equal zero for some value of. Most often, a zero will occur at c when the linear factor ( c) occurs in both the numerator and denominator of a rational function. This factor cancels and results in a domain restriction for the simplified form of R. The long run behavior and horizontal asmptote of R can be determined b the ratio of leading terms of p and q. Determining the horizontal asmptote of a rational function: Let m be the degree of the function p in the numerator and let n be the degree of the function q in the denominator. If m < n, then the horizontal asmptote is = 0. If m = n, then the horizontal asmptote is = c where c is a real number determined b the ratio of leading coefficients. If m > n, then no horiztonal asmptote eists. If m = n + 1, then an oblique asmptote eists. Eample 1. Determine the long-run behavior of the following functions. (a) R() = 3 In the long run, R() 3. Thus R looks like the function f where f() = 3 in the long run. Horizontal Asmptote: = 0 As, R() 0 As, R() 0 1

(b) R() = 3 5 + 6 In the long run, R() 3. Thus R looks like the function f where f() = 3 As, R() 0 As, R() 0 Horizontal Asmptote: = 0 in the long run. (c) R() = 5 6 + 1 In the long run, R(). Thus R looks like the function f where f() = 1 in the long run. Horizontal Asmptote: = 1 As, R() 1 As, R() 1 (d) R() = + 3 In the long run, R(). Thus R looks like the function f where f() = in the long run. As, R() As, R() No horizontal asmptote. This function has an oblique asmptote. (e) R() = 3 7 1 In the long run, R() 7. Thus R looks like the function f where f() = 1 5 in the long run. As, R() As, R() No horizontal or oblique asmptote. Eample. Determine an horiztonal asmptotes for the following rational functions. (a) R() = 5 + 1 10 + 6 Since the degree of the numerator is 1 and the degree of the denominator is, the horizontal asmptote is = 0. (b) R() = 5 + 1 10 + 6 Since the degrees of the numerator and denominator are, the horizontal asmptote is = 5 10, which simplifies to = 1. (c) R() = 53 + 1 10 + 6 Since the degree of the numerator is 3 and the degree of the denominator is, no horizontal asmptote eists. An oblique asmptote eists. (d) R() = 5 + 1 10 + 6 Since the degree of the numerator is and the degree of the denominator is, no horizontal or oblique asmptote eists. Instructor: A.E.Car Page of 16

Eample 3. Graph the rational function R() = + 3 b completing the following: factor and simplif 3 R(); state the domain; determine an holes and/or vertical asmptotes; determine the long-run behavior and an horiztonal asmptotes; determine an horizontal and vertical intercepts; determine the zeros and their multiplicit. Simplifing R(): R() = + 3 3 ( 3)( 1) = 3 = 1, 3 Thus R() is a linear function with a restricted domain. The domain of R is { 3} or (, 3) (3, ). There is a hole at 3. To determine the output of this hole, we use the simplified form of R: R(3) 3 1 = Thus the point (3, ) will be graphed as an open circle. There are no vertical asmptotes. No horizontal asmptote eists. In the long run, R() looks like the function defined b f() =. As, As, R() R() As R(0) = 1, the vertical intercept is (0, 1). The horizontal intercepts (and zeros) of R occur where R() = 0: The horizontal intercept is (1, 0). R() = 0 1 = 0 = 1 The zero of R is 1 with multiplicit 1, so the graph crosses the horizontal ais at 1. = +3 3 6 Figure 1 6 6 6 Instructor: A.E.Car Page 3 of 16

Eample. Graph the rational function R() = 3 6 b completing the following: factor and simplif + 6 R(); state the domain; determine an holes and/or vertical asmptotes; determine the long-run behavior and an horiztonal asmptotes; determine an horizontal and vertical intercepts; determine the zeros and their multiplicit. Simplifing R(): The domain of R is { 3, }. R() = 3 6 + 6 3( ) = ( + 3)( ) = 3 + 3, There is a hole at. To determine the output of this hole, we use the simplified form of R: R() 3 + 3 = 3 5 Thus the point (, 3 5) will be graphed as an open circle. The vertical asmptote is = 3. Since the degree of the numerator is less than the degree of the denominator, the horizontal asmptote is = 0. In the long run, R() 3 = 3 : As, R() 0 As, R() 0 As R(0) = 1, the vertical intercept is (0, 1). The horizontal intercepts (and zeros) of R occur where R() = 0: There is no horizontal intercept. There are no zeros of R. R() = 0 3 0 Instructor: A.E.Car Page of 16

Table 1. Behavior of R Zeros/ VA Interval (, 3) ( 3, ) - - R() R( ) = 3 R( ) = 3 +/ - (below) + (above) Point (, 3) (, 3) = -3 Figure 6 = 3 6 + 6 = 0 6 6 6 Instructor: A.E.Car Page 5 of 16

Eample 5. Graph the rational function R() = 5 6 b completing the following: factor and simplif + 1 R(); state the domain; determine an holes and/or vertical asmptotes; determine the long-run behavior and an horiztonal asmptotes; determine an horizontal and vertical intercepts; determine the zeros and their multiplicit. Simplifing R(): R() = 5 6 + 1 ( 6)( + 1) = ( + )( 3) The domain of R is {, 3}. There are no holes. The vertical asmptotes are = and = 3. Since the degree of the numerator is the same as the degree of the denominator, the horizontal asmptote is determined b the ratio of leading coefficients: = 1. To determine if the function crosses its horizontal asmpote, we solve R() = 1: R() = 1 5 6 + 1 = 1 5 6 = + 1 6 = 6 = 1 The graph of = R() crosses the horizontal asmptote at the point (1, 1). In the long run, R() = 1: As, R() 1 As, R() 1 As R(0) = 1, the vertical intercept is ( 0, 1 ). The horizontal intercepts (and zeros) of R occur where R() = 0: R() = 0 ( 6)( + 1) = 0 6 = 0 or + 1 = 0 = 6 or = 1 The horizontal intercepts are (6, 0) and ( 1, 0). The zeros of R are -1 and 6, as found in the previous step. Each has a multiplicit of 1, so the graph crosses the horizontal ais at -1 and 6. Instructor: A.E.Car Page 6 of 16

Table. Behavior of R Zeros/ VA Interval (, ) (, 1) ( 1, 3) (3, 6) (6, ) -6-1 8 R() R( 6) = 10 3 R( ) = 5 R(1) = 1 R() = 5 R(8) = 3 10 +/ + (above) - (below) + (above) - (below) + (above) ( ) ( ) ( ) ( ) Point 6, 10 3, 5 (0, 1), 5 8, 3 10 Figure 3 6 = 5 6 + 1 = 1 6 6 6 = - = 3 Instructor: A.E.Car Page 7 of 16

Eample 6. Graph the rational function R() = 8 b completing the following: factor and simplif R(); state the domain; determine an holes and/or vertical asmptotes; determine the long-run behavior and an horiztonal asmptotes; determine an horizontal and vertical intercepts; determine the zeros and their multiplicit. Simplifing R(): R() = 8 8 = ( )( + ) The function is full simplified. The domain of R is {, }. There are no holes. The vertical asmptotes are = and =. Since the degree of the numerator is less than the degree of the denominator, the horizontal asmptote is = 0. To determine if the graph of = R() crosses this horizontal asmptote, we solve R() = 0: In the long run, R() 8 : R() = 0 8 = 0 8 0 As, R() 0 As, R() 0 As R(0) =, the vertical intercept is (0, ). The horizontal intercepts (and zeros) of R occur where R() = 0: There are no horizontal intercepts. There are no zeros of R. R() = 0 8 0 Instructor: A.E.Car Page 8 of 16

Table 3. Behavior of R Zeros/ VA Interval (, ) (, ) (, ) - 0 R() R( ) = 3 R(0) = R() = 3 +/ + (above) - (below) + (above) ( ) ( ) Point, 3 (0, ), 3 Figure 6 = 8 = 0 6 6 6 = - = Instructor: A.E.Car Page 9 of 16

Eample 7. Graph the rational function R() = 8 b completing the following: factor and simplif 9 R(); state the domain; determine an holes and/or vertical asmptotes; determine the long-run behavior and an horiztonal asmptotes; determine an horizontal and vertical intercepts; determine the zeros and their multiplicit. Simplifing R(): R() = 8 9 ( )( + ) = ( 3)( + 3) The domain of R is { 3, 3}. There are no holes. The vertical asmptotes are = 3 and = 3. Since the degree of the numerator is the same as the degree of the denominator, the horizontal asmptote is determined b the ratio of leading coefficients: = 1 1 = 1. To determine if the function crosses its horizontal asmpote, we solve R() = 1: R() = 1 8 = 1 9 8 = 9 = 1 = 1 The graph of = R() crosses the horizontal asmptote at the point ( 1, 1). In the long run, R() = 1: As, R() 1 As, R() 1 As R(0) = 8 9, the vertical intercept is ( 0, 8 9). The horizontal intercepts (and zeros) of R occur where R() = 0: R() = 0 ( )( + ) = 0 = 0 or + = 0 = or = The horizontal intercepts are (, 0) and (, 0). The zeros of R are - and, as found in the previous step. Each has a multiplicit of 1, so the graph crosses the horizontal ais at - and. Instructor: A.E.Car Page 10 of 16

Table. Behavior of R Zeros/ VA Interval (, 3) ( 3, ) (, 3) (3, ) (, ) - -.5 0.5 3.5 6 R() R( ).3 R(.5) 1. R(0.5) = 1 R(3.5) 0.85 R(6) 0.6 +/ + (above) - (below) + (above) - (below) + (above) Point (,.3) (.5, 1.) (0.5, 1) (3.5, 0.85) (6, 0.6) **Function values approimated with a calculator.** Figure 5 6 = 8 9 = 1 10 8 6 6 8 10 6 = -3 = 3 Instructor: A.E.Car Page 11 of 16

How to find a possible formula for a rational function: State an zeros. Use these to determine factors and the multiplicit of each factor that appears in the numerator. State an vertical asmptotes. Use these to determine factors and the multiplicit of each factor that appears in the denominator. If a hole appears at = a, then put the factor ( a) in both the numerator and denominator. Use one other point to determine if there is a constant factor other than 1. Eample 8. Find a possible formula for the rational function graphed in Figure 6 below. Figure 6 6 = ( ) 0, 8 3 6 6 6 = 1 = 3 Table 5. Behavior of R Zeros/ VA Multiplicit Factor Numerator/denominator zero: even () ( ) numerator VA: = 1 odd (1) ( 1) denominator VA: = 3 odd (1) ( 3) denominator Thus a possible formula for R() is of the form R() = k( ) ( 1)( 3) Instructor: A.E.Car Page 1 of 16

where k is a constant factor. To determine k, we will use one point on the graph that is not a horizontal intercept. Here, the point ( 0, 8 3) is given: R(0) = 8 3 k(0 ) (0 1)(0 3) = 8 3 3 k = 8 3 k = The constant factor of k can also be determined based upon the horizontal asmptote =. A possible formula for R() is R() = ( ) ( 1)( 3) Instructor: A.E.Car Page 13 of 16

Eample 9. Find a possible formula for the rational function graphed in Figure 7 below. Figure 7 6 (0, 1) = 0 6 6 6 = - = Table 6. Behavior of R Zeros/ VA Multiplicit Factor Numerator/denominator VA: = odd (1) ( + ) denominator VA: = even () ( ) denominator Thus a possible formula for R() is of the form R() = k ( + )( ) where k is a constant factor. To determine k, we will use one point on the graph that is not a horizontal intercept. Here, the point (0, 1) is given: R(0) = 1 k (0 + )(0 ) = 1 k 16 = 1 k = 16 Therefore a possible formula for R() is R() = 16 ( + )( ) Instructor: A.E.Car Page 1 of 16

Eample 10. Oblique Asmptotes The graph of R() = 5 8 looks like the function defined b f() = 1 in the long run. We know that this function has an oblique asmptote as the degree of the numerator is greater than the degree of the denominator. To determine the equation of the oblique asmptote, either polnomial long division or a graphing calculator are needed. Using the epand ke on a graphing calculator to perform polnomial long division, we find: We use this to write: R() = 5 8 = 9 8 + 1 1 The first term in the epanded R() is the remainder of this long division. The epression 1 1 is used to determine the equation of the oblique asmptote, which is = 1 1. The function and its oblique asmptote are graphed in Figure 8 below. = 1 1 = 5 8 Figure 8 6 6 6 6 = Instructor: A.E.Car Page 15 of 16

Eample 11. Other Improper Rational Functions The graph of R() = 3 7 1 looks like the function defined b f() = 1 5 in the long run. Graphs of = R() and = f() are shown in Figure 9 below. Figure 9 30 0 10 3 1 1 3 10 0 = 1 5 = 3 7 1 = -0.5 30 = 0.5 Instructor: A.E.Car Page 16 of 16