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Question Number 123199 by aurpeyz last updated on 23/Nov/20

sketch the curve of y=((x^2 −1)/(x^2 −7x−12))

sketchthecurveofy=x21x27x12

Answered by MJS_new last updated on 24/Nov/20

y=((x^2 −1)/(x^2 +7x−12))  y′=−((7x^2 +22x+7)/((x^2 +7x−12)^2 ))  y′′=((2(7x^3 +33x^2 +21x+83))/((x^2 −7x−12)^3 ))  y=0 ⇒ zeros at x=±1  x^2 +7x−12=0 ⇒ asymptotes at x=((7±(√(97)))/2) (≈−1.42 and 8.42)  y=1+((7x+11)/(x^2 −7x−12)) ⇒ asymptote at y=1 (for x=±∞)  y′=0 ⇒ extremes at x=−((11±6(√2))/7) (≈−2.78 and −.36)  y′′  { ((>0 for x=−((11+6(√2))/7) ⇒ local min at ≈ (((−2.78)),((.44)) ))),((<0 for x=−((11−6(√2))/7) ⇒ local max at ≈ (((−.36)),((.09)) ))) :}  y′′=0 ⇒ inflection point ≈ (((−4.62)),((.49)) )  y′′  { ((<0 for x<≈−4.62 ⇒ curvature −)),((>0 for ≈−4.62<x<≈−1.42 ⇒ curvature +)),((<0 for ≈−1.42<x<≈8.42 ⇒ curvature −)),((>0 for x>≈8.42 ⇒ curvature + )) :}    now sketch it

y=x21x2+7x12y=7x2+22x+7(x2+7x12)2y=2(7x3+33x2+21x+83)(x27x12)3y=0zerosatx=±1x2+7x12=0asymptotesatx=7±972(1.42and8.42)y=1+7x+11x27x12asymptoteaty=1(forx=±)y=0extremesatx=11±627(2.78and.36)y{>0forx=11+627localminat(2.78.44)<0forx=11627localmaxat(.36.09)y=0inflectionpoint(4.62.49)y{<0forx<≈4.62curvature>0for4.62<x<≈1.42curvature+<0for1.42<x<≈8.42curvature>0forx>≈8.42curvature+nowsketchit

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