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Question Number 123199 by aurpeyz last updated on 23/Nov/20
sketchthecurveofy=x2−1x2−7x−12
Answered by MJS_new last updated on 24/Nov/20
y=x2−1x2+7x−12y′=−7x2+22x+7(x2+7x−12)2y″=2(7x3+33x2+21x+83)(x2−7x−12)3y=0⇒zerosatx=±1x2+7x−12=0⇒asymptotesatx=7±972(≈−1.42and8.42)y=1+7x+11x2−7x−12⇒asymptoteaty=1(forx=±∞)y′=0⇒extremesatx=−11±627(≈−2.78and−.36)y″{>0forx=−11+627⇒localminat≈(−2.78.44)<0forx=−11−627⇒localmaxat≈(−.36.09)y″=0⇒inflectionpoint≈(−4.62.49)y″{<0forx<≈−4.62⇒curvature−>0for≈−4.62<x<≈−1.42⇒curvature+<0for≈−1.42<x<≈8.42⇒curvature−>0forx>≈8.42⇒curvature+nowsketchit
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