Rational Functions. Rational functions are the ratio of two polynomial functions. Qx bx b x bx b. x x x. ( x) ( ) ( ) ( ) and


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1 Rtionl Functions Rtionl unctions re the rtio o two polynomil unctions. They cn be written in expnded orm s ( ( P x x + x + + x+ Qx bx b x bx b n n 1 n n m m 1 m + m m + 0 Exmples o rtionl unctions in expnded orm re 3x 5x + 7 nd 3 5x 3x + x 4 3 8x + 4x 3 Rtionl unctions cn lso be written in ctored orm s m p1 p ( 1 ( ( j n x c x c x c b x d x d x d q1 q ( ( ( 1 p + p + + p n, q + q + + q m 1 j 1 where the c's nd d's re rel or complex numbers. Exmples o rtionl unctions in ctored orm re ( 1( + 1 x x x ( x 3 ( x+ 4 nd k k p q j k ( x 3( x+ 4 ( 5x 1( x The domin is ll rel numbers except where the is zero.. All inormtion regrding roots or zeros o the unction come rom the. 3. The number o rel or complex zeros is the o the numertor. x x 4. Assume tht is rel number nd is ctor o the numertor... x. The grph will cross the xxis nd chnge sides t i the exponent on the ctor is. x b. The grph will touch the xxis but sty on the sme side t i the exponent on the ctor is.
2 5. All inormtion regrding verticl symptotes o the unction come rom the. 6. As x pproches verticl symptote, the y pproches or. x x 7. Assume tht is rel number nd is ctor o the denomintor.... The grph will be symptotic in opposite directions to verticl line t i the exponent on the ctor is. x b. The grph will be symptotic in the sme direction to verticl line t i the exponent on the ctor is. x 8. The only plces tht rtionl unction cn chnge rom positive to negtive or negtive to positive re t n or.. The rtionl unction will chnge signs nd hve grph on dierent sides o the xxis only i the exponent on the ctor is. b. The rtionl unction will sty the sme sign nd hve grph on the sme side o the xxis only i the exponent on the ctor is. Prctice: Determine which vlues o x re xintercepts nd which re verticl symptotes. For ech x vlue, determine whether the unction will chnge signs or sty the sme sign on either side o these vlues. Function xintercepts verticl symptotes chnge sme chnge sme ( x+ ( x 3 ( x+ 4 ( x ( 4( + 4 x x x ( x 5( x+ 3 4
3 9. The right nd let hnd behvior o the grph re determined by the degrees o the numertor nd denomintor.. There will be no horizontl symptote i there re more x s in the. b. The grph will hve horizontl symptote t y0 (the xxis i there re more x s in the. c. The grph will hve horizontl symptote t the rtio o the leding coeicients i the degree in the numertor nd the denomintor re the. d. The grph will hve n oblique (slnt symptote i there is exctly one more x in the. The slope o the oblique symptote is the o the leding coeicients. ( (b (c (d
4 10. Horizontl symptotes re only symptotes s x pproches or ; tht is, to the r or the r o the grph. 11. The unction my cross the horizontl symptote in the middle section o the grph. Bsiclly, the middle section is nything between the smllest nd lrgest or. 1. Although the grph my cross horizontl symptote, the grph cn never cross. 13. As you grph the unction, strt on the nd work your wy to the. 14. I there is common ctor between the numertor nd the denomintor, then there my or my not be verticl symptote t tht vlue. Assume tht x is common ctor (the powers my be dierent o both the numertor nd denomintor. x. There will be verticl symptote t i, ter simpliying the rtionl unction, is still in the. x x b. There will be hole in the grph on the xxis t i, ter simpliying the rtionl unction, x is still in the. The vlue x ( is / is not n xintercept. x c. There will be hole in the grph, but not on the xxis, t i, ter simpliying the rtionl unction, x is no longer in the numertor or the denomintor. x d. Regrdless o where the ends up ter simpliiction, it is importnt to note tht x is not in the o the unction becuse it originlly cused division by. Prctice: Identiy the x vlue ssocited with the common ctor nd determine i there will be n xintercept, hole on the xxis, hole o the xxis, or verticl symptote there. ( x ( x+ ( x+ ( x ( x+ ( x+ ( x+ ( x
5 15. The yintercept is the rtio o the o the numertor nd denomintor. 16. Rtionl unctions re, there re no shrp turns. 17. Rtionl unctions re, they cn be drwn without liting up your pencil, except t or. 18. Complex roots ren't or nd the unction cnnot chnge signs t them. 19. Complex roots my need to be dded so tht the right nd let hnd behvior is correct. While it my not lwys give the correct grph, the simplest complex ctor to insert into the rtionl unction is. 0. Complex roots my lso be necessry when there re extr in the grph, just like they were or polynomil unctions. 1. In the cse where there is no horizontl symptote, the right hnd behvior (s x + is determined by looking t the o the leding coeicients. Be sure to py ttention to whether or not there is leding negtive sign.. Even i there is horizontl or oblique symptote, you cn still determine whether the right hnd side is or by looking t the sign o the rtio o the leding coeicients. 3. The let hnd behvior (s x is similr to tht o polynomil unctions, except tht insted o looking t the degree o the polynomil, you should look t the in the degrees o the numertor nd denomintor.. I the dierence o the degrees is, then the grph will be on the sme side o the horizontl symptote s the right side. b. I the dierence o the degrees is, then the grph will be on opposite sides o the horizontl symptote s the right side. 4. For lrge (postive or negtive vlues o x, only the mtter. All other powers o x re insigniicnt in comprison.
6 5. Answer the questions bout the rtionl unction (be sure to note the  in ront. 3 8x ( x ( x+ 4 ( x 5 3 ( x ( x+ ( x ( x Wht is the domin o the unction? b. Simpliy the unction, stting ny necessry restrictions. c. Check the box in ech row tht describes the grph o the unction t the indicted vlue o x. vlue hole, not on xxis hole on x xis verticl symptote regulr x intercept none o these x 5 x x 3 x 1 x 4 d. Which o the ollowing sttements describes the grph o the unction? (Circle one nd ill in the blnk i pproprite i. There is no horizontl symptote. ii. There is n oblique symptote with slope m. iii. The horizontl symptote is the xxis (y 0. iv. There is horizontl symptote t the line y. e. Mke sign chrt or the unction.. Sketch the grph o the unction.
7 6. Answer the questions bout the rtionl unction. ( x ( x ( x ( x. Wht is the domin o the unction? b. Simpliy the unction, stting ny necessry restrictions. c. Check the box in ech row tht describes the grph o the unction t the indicted vlue o x. vlue x 1 x x 3 x 3 x 4 hole, not on xxis hole on x xis verticl symptote regulr x intercept none o these d. Which o the ollowing sttements describes the grph o the unction? (Circle one nd ill in the blnk i pproprite i. There is no horizontl symptote. ii. There is n oblique symptote with slope m. iii. The horizontl symptote is the xxis (y 0. iv. There is horizontl symptote t the line y. e. Mke sign chrt or the unction.. Sketch the grph o the unction.
8 7. Write unction whose grph is shown.. b. c.
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