Rational Functions Deinition A rational unction can be written in the orm () N() where N() and D() are D() polynomials and D() is not the zero polynomial. *To ind the domain o a rational unction we must look at what values make the denominator = 0. These numbers must be ecluded rom the domain. Eample: Find the domain o 1. 9 Set 9 = 0 and solve. ( 3)( + 3) = 0 = 3 or = -3 We say the domain is all real numbers ecept 3 and -3. We can also list it as (-, -3) (-3, 3) (3, ). 1
Look at 35. Questions: What happens to y as approaches rom the let? y approaches - What happens to y as approaches rom the right? y approaches + What happens to y as approaches? y approaches 3 What happens to y as approaches -? y approaches 3
Notation: () - as - () as + () 3 as - () 3 as *The line = is called the vertical asymptote. *The line y = 3 is called the horizontal asymptote. Deinition o asymptotes: 1. The line = a is a vertical asymptote o the graph o i () ± as a, either rom the right or rom the let.. The line y = b is a horizontal asymptote o the graph o i () b as ±. Look at: ( ) 1 1 The horizontal asymptote is y = The vertical asymptote is = -1 3
Look at ( ) 4 1 The horizontal asymptote is y = 0 There is no vertical asymptote. *Note that the graph crosses the asymptote at the origin. Sometimes the graph o a rational unction will cross asymptotes at or around the origin. Look at ( ) ( 1) The horizontal asymptote is y = 0 The vertical asymptote is = 1 4
Look at ( ) There is no horizontal asymptote. The vertical asymptote is = Rules or Asymptotes o Rational Functions: Let be a rational unction given by ( ) N( ) D( ) a b n m n m a b n1 m1 n1 n1... a1 a... b b 1 0 0 where N() and D() have no common actors. 1. The graph o has vertical asymptotes at the zeros o the polynomial D(). (ie. where D() = 0) 5
. The graph o has one or no horizontal asymptote, depending on the degree o N and D. a. I n < m, then y = 0 is the horizontal asymptote o the graph o. b. I n = m, then y = the graph o. a b n m is the horizontal asymptote o c. I n > m, then the graph o has no horizontal asymptote. Eamples: Find the horizontal and vertical asymptotes. 5 a) 4 6 horizontal asymptote: y = 1 vertical asymptote: = 3 6
1 b) horizontal asymptote: y = 0 vertical asymptote: = c) 1 horizontal asymptote: none vertical asymptote: = -1 1 d) 6 ( ) 1 ( 3)( ) horizontal asymptote: y=0 vertical asymptote: = 3 and = - 7
1 Look at: 1 What is the domain? all real numbers ecept 1 and -1 Look at the graph on your calculator: Is there an asymptote at = -1? No 8
On your calculator, look at the table values or 1 and -1. [TblSet] TblStart = - and the [Table] What values are given or 1 and -1? Both show error. 1 Look at 1 again. Factor it to ( ) ( 1 1)( 1) 1 1 1 Graph 1 and you will get the same graph. 1 **Since 1 is our original equation, we cannot ignore our original domain, even i the equation simpliies to something else. Because -1, we will put a hole at = -1. 9
1 The graph o 1 should look like: There is a hole at = -1. 10
Guidelines or Analyzing Graphs o Rational Functions: CHAT Pre-Calculus N() Let () = D() where N() and D() are polynomials with no common actors. 1) Find and plot the y-intercept (i any) by evaluating (0). ) Find the zeros o the numerator (i any) by solving the equation N() = 0. Then plot the corresponding -intercepts. 3) Find the zeros o the denominator (i any) by solving the equation D() = 0. Then sketch the corresponding vertical asymptotes. 4) Find and sketch the horizontal asymptote (i any) by using the rule or inding the horizontal asymptote o a rational unction. 5) Test or symmetry. 6) Plot at least one point between and one point beyond each -intercept and vertical asymptote. 7) Use smooth curves to complete the graph between and beyond the vertical asymptotes. 11
Note: Because the unction can only change signs at its zeros and vertical asymptotes, we use these values to determine test intervals. Eample: Sketch ( ) 3 4. y-intercept: (0,0) -intercept: (0,0) vertical asymptote: = -4 horizontal asymptote: y = 3 Points in test intervals: (-5, 15) (-, -3) (, 1) 1
Eample: Sketch. CHAT Pre-Calculus y-intercept: (0,0) -intercept: (0,0) vertical asymptote: =, = -1 horizontal asymptote: y = 0 Points in test intervals: (-3, -0.3) (-0.5, 0.4) (1, -0.5) (3, 0.75) 13
Eample: Sketch ( ) ( 9) 4 ( ) ( 9) 4 ( 3)( 3) ( )( ) y-intercept: (0,4.5) -intercept: (-3,0) and (3,0) vertical asymptote: =, = - horizontal asymptote: y = symmetry: y-ais because it is an even unction Points in test intervals: (-6, 1.69) (-.5, -.44) (0.5, 4.67) (.5, -.44) (6, 1.69) 14
Look at: ( ) 1 *This unction has a slant asymptote. This happens only i the degree o the numerator is eactly one more than the degree o the denominator. To Find Slant Asymptotes: Use long division to divide the denominator into the numerator. The equation o the asymptote is the quotient, ecluding the remainder. 15
Look again at. 1 Do long division: So we have: 1 0 + - + 0 - - 1 1 Then the slant asymptote is (or y = - ) Eample: Sketch y-intercept: (0,0) -intercept: (0,0) vertical asymptote: = slant asymptote: y=+ Points: (-1/, -0.1) (1, -1) (3, 9) 16
Rules or Asymptotes o Rational Functions: Let be a rational unction given by ( ) N( ) D( ) a b n m n m a b n1 m1 n1 n1... a1 a... b b 1 0 0 where N() and D() have no common actors. 3. The graph o has vertical asymptotes at the zeros o the polynomial D(). (ie. where D() = 0) 4. The graph o has one or no horizontal asymptote, depending on the degree o N and D. a. I n < m, then y = 0 is the horizontal asymptote o the graph o. b. I n = m, then y = the graph o. a b n m is the horizontal asymptote o c. I n > m, then the graph o has no horizontal asymptote. 5. I n=m+1, then the graph o has a slant asymptote at y=q(), where q() is the quotient obtained rom the division algorithm, ecluding any remainder. 17