4.5 Rational functions.

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1 4.5 Rational functions. We have studied graphs of polynomials and we understand the graphical significance of the zeros of the polynomial and their multiplicities. Now we are ready to etend these eplorations to rational functions. The eamples below contain a sequence of equations and their graphs. There would seem to be three ways to use these as interactive classroom eplorations. My favorite is the first: give the students the graphs and ask for the equations give the students the equations and ask for the graphs give them both and get them to talk about the relationship In talking about the graphs the key new idea of course will be the asymptotes, vertical, horizontal and slant and how they are approached from each side. The vertical asymptotes have the same relationship to the zeros in the denominator as the intersections with the -ais have to the zeros in the numerator. The horizontal or slant asymptotes have to do with the difference in degree between the numerator and the denominator. Eample. -intercepts: simple zero in the numerator at =. vertical asymptotes: simple zero in the denominator at =. horizontal asymptote: What happens as gets large positive and negative? Divide top and bottom by : / / as approaches ±, this approaches. So the horizontal asymptote is. Eample. ( )( ) -intercepts: simple zero in the numerator at =. vertical asymptotes: simple zeros in the denominator at = and =. horizontal asymptote: Rewrite: and divide top and bottom by : + / / as approaches ±, this clearly approaches 0. So the graph is asymptotic to the -ais. Rational functions 9/3/007

2 Eample 3. ( + ) ( )( ) -intercepts: simple zeros in the numerator at = and =0. vertical asymptotes: simple zeros in the denominator at = and =. horizontal asymptote: Rewrite: + and divide top and bottom by : + / / as approaches ±, this clearly approaches. So the horizontal asymptote is. Eample 4. ( + ) ( ) ( ) -intercepts: simple zero on top at =0, and a double zero at =. vertical asymptotes: simple zero on the bottom at =, double zero at =. Note carefully the effect of this the epression does not change sign at =, so the curve goes up the asymptote from both sides. horizontal asymptote: Instead of rewriting note that the leading term in both the top and the bottom will be 3. If we divide top and bottom by 3 both will have the form: + (terms which approach zero for large. Thus the horizontal asymptote is. At simple (i.e. single) zeros both in the numerator and the denominator y changes sign. At double zeros, like the denominator zero at =, it does not. Rational functions 9/3/007

3 Eample 5. ( + ) ( ) -intercepts: simple zero on top at =0, and double zero at =. vertical asymptotes: double zero in the bottom at =. The curve goes down the asymptote from both sides. slant asymptote: The degree of the top is one more than the degree of the bottom, so for large we epect the behaviour to look like a degree polynomial, i.e. a line. But what line should it be? Here s a quick and dirty answer. Epand the top and bottom: Divide top and bottom by : /. + / / Letting get large many terms approach zero, leaving us with = Thus the asymptote must be +4. This argument certainly seems reasonable, but it gives the wrong answer. What s wrong is that it doesn t actually show what needs to be shown, that the distance between the curve and the line +4 approaches zero as gets large. [And it doesn t!] Here s a proper argument. Use long division to divide the denominator + + into the numerator You get a quotient of + and a remainder of. Thus: = ( + ) Rewrite: ( + ) = + + and this tells us that the vertical distance between and (+) + approaches zero as get large. The asymptote is the line +. This argument seems so reasonable, it s hard to see what s wrong with it. But it s wrong. At the end of the day it doesn t prove the one fact that needs to be established, that ( + 4) + approaches zero as gets large. Take some comfort from the fact that when I was young I found this argument totally acceptable. I well remember my dismay upon discovering that it was fallacious. + Rational functions 9/3/007 3

4 Eample 6. Find the equation I used to draw this rational function. What sort of asymptote does it have as gets large? -intercepts: simple zeros at = 3,,,. So we have factors (+3), (+), ( ) and ( ) in the numerator. vertical asymptotes: Asymptote at =0 and the function does not change sign here (it goes up the asymptote from both sides). Thus we have an even power of in the denominator likely. Try the form: ( + 3)( )( ) The numerator has degree 4 and the denominator has degree, and the difference is. Thus the asymptote (as gets large) will be a polynomial of degree so a parabola! To find it, divide the denominator into the numerator: ( + 3)( )( ) ( = 4 3 )( = ) + = (. + 6) The asymptote is the parabola + 7, drawn at the right. You can do some simple numerical checks. Problems. Find a rational function that has zeros at = 0 and 3 (and perhaps others), vertical asymptotes are = ±, and as gets large, is asymptotic to.. Below you will find a number of sets of graphs together with the defining equation. In each case discuss the correspondence between them. Relate the geometry to the algebra the asymptotes and the intersections with the ais to the terms in the numerator and the denominator of the equation. Sometimes you will observe a geometric symmetry. Is there a corresponding symmetry in the equation? Rational functions 9/3/007 4

5 Set (a) 3 (b) 3 (c) ( 3) (d) ( )( ) ( 3) Rational functions 9/3/007 5

6 Set (a) ( ) ( 4) (b) ( ) ( 4) (c) ( ) + ( 4) Rational functions 9/3/007 6

7 Set 3 (a) ( ) (b) ( ) (c) ( ) (d) ( ) Rational functions 9/3/007 7

8 3. Find rational equations for each of the following graphs. (a) (b) (c) (d) Rational functions 9/3/007 8

9 (e) (f) (g) (h) Rational functions 9/3/007 9

10 4. Rewrite the rational epressions below in a way that will allow you to easily draw their graphs. (a) (b) (c) 3 (d) 3 3 Rational functions 9/3/007 0

11 (e) + (f) Rational functions 9/3/007

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