Section%203.5

# Section%203.5 - MA 180 Precaloulus Professor Terry Section...

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Unformatted text preview: MA 180 - Precaloulus Professor Terry Section 3.5: Rational Functions E Rational Functions and Their Domains Afunction f is a rational function if f aha—33%, where g(x) and h(x) are polynomials. x The domain of f consists of all real numbers except the S of the denominator h(x). Sketch the graph of f and ﬁnd its domain. 2—4 100:" x2+1 Domain KER x'Z-H 7 O —3 f(x)-x_3 f(x)=2x_+l_ x +2x—3 Domain FR QLQ/W X:[ X:_3 X2‘t2x‘« 3430 (x43) (5H) 40 X15540 Qt: X=Hr0 Xi—zb , 31% l MA180 SecS.5 Fall03 ewt 2 —3x—-4 ftx)=3‘——-—— x2+x-—6 Domain’Q excﬁy’é A12 ZQXJFHS X 2% MAL; ﬁfe? 4’ '0“ 53002) 4 o x+%=#o gig )‘vaia Xirﬁ x4e _2x2—x—3 “‘7 Domain excgp¥ X: 3 Li 7 * 7 . ' . ‘ x 2 ‘9' df Vertical and Horizontal As m totes CYCL/QCcL/QMD A X1402; _.____ .7 C0 Poem/cw x .3 5 X ‘ 3 * Question 1 What happens to the function values of f (x) when x is close to (but not equal to) a zero of the denominator? Question 2 What happens to the function values of f (x) when x is large positive or when x is large negative? Deﬁnition of Vertical Asymptotes: The lin:- x = a s a vertical asymptote for the graph of a function flf ¢___..__.__—~ f(x)—>oo or f(x)—>-oo as x approaches a from either the left or the right. MA180 Sec3.5 Falloa ewt Find the equation of the vertical asymptotes. f(x)= ‘3 X2 E) x-3 f(x):x2—4 Vex/{164,0 j x2+1 ' ‘2; . ., ._ X + 1 cl: '0 +1 “mix: X‘Z'tZX‘E’stgL, l “2 ', lg f(x)=x2_‘§3‘_‘i ){%rt- X—(g=o f(x)=2x2"‘"3 X’~& “—o H M: a ﬁg“ _____ __ %_ f(x):x2_9 z @a _ X+ @X'h (“GLC (Esrng x_3 M 3 (Due/WE t‘ﬁ‘ Deﬁnition of Horizontal As m totes: The “an is a horizontal asymptote for the graph of a function f if f(x)-)c as x—>oo ores x—9—oo. n—l k—l n anx + an_1x k +'"+alx+a0 ,where an :20 and +...+b1x+b0 Theorem on Horizontal Asymptotes: Let f (x): bkx + bk_1x bk ¢ 0. charm; M 4 D (1) If (n <_k) then the x-axis (y = 0) is the horizontal asymptote of V the graph of f . E- A 4 I (2) HQ) then the line y e an wk (the ratio of leading N10 4 rig/ta Al e cicﬁbae i3 (ageincients) is the horizontal asymptote of the graph of f . (3) [When the graph of f has no horizontal asymptote. / instead, either f(x)—)oo or f(x)—->—oo as - x—)oo oras x—~>—00. _ MA180\$ec3.5Fall03ewt j: g (£17 (KW M > at farm“ "D W a aﬂjk‘fﬂaff' Find the equation of the horizontal asymptote. ftx)=x“_33 4‘3 ''''' [W 2 f(x)=x ‘4 M .x2‘+1 . I ' at 2 "T 1 I 4w M <- dw a f(x)=":'3"‘4 “Le‘tw M x +x—6 L6 2x2—x—3 €443 “:91, Mo f(x)= x—2 E1173 RH: A 2- r > '2 7 D “30139 2 m a ’55 if“ _ . Hm a} x a 5 @_9w_*ﬁf¥j_j J Guidelines for Sketchin the Gra hof a Rational Function Page 255 in your text. Sketch the graph of f. f(x)= ﬁx” ‘ x +2x—3 X ‘- t m/ifoir 5 x+ { 2-. o M WM X: "'I g‘ 1 “Jim )6 .-_ O ¥CO ]: f‘ -_—- z “"1 9‘4 2 0 3 3 ‘4; X' 2'121— 5'1 O (Dementnmgm/D (X4 '5')(X-l) 7— O Xz—B y: I {id-Er CJ~€C(L(1 104-4—6ch b e... (3:0 '35) m WM Ho \$4.44.? X‘f l 20 MA‘IBO Sec3.5 Fa|103 ewt X 2 -— i x2—3x-4 x2+x—6 Sketch the graph of f . f (x) = Obiigue Asymptotes Since the degree of the numerator is greater than the degree of the denominator; there is M 0 horizontal asymptote. But, since the degree of the numerator is greater than the degree of the denominator by 1 there is an oblique asymptote. Use long division to ﬁnd f (x) = : (ax +b) + % where either r(x) = 0 or the degree of r(x) is x 3: less than the degree of h(x). using the theorem on horizontal asymptotes (1), 71?.) o as x _, m or x —) —oo. And, y = ax + b is the oblique asymptote. J: Find the oblique asymptote and sketch the graph of f . Q x +3 X-ch Qx2*X'-3 - @XZ “ LIX) 534—5 __(3x‘- (a) 5 MA1 80 5603.5 FallOB ewt ...
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Section%203.5 - MA 180 Precaloulus Professor Terry Section...

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