Question #315801

Find the asymptotes of y^2(a+x) =x^2(3a-x) parallel to coordinate axes



1
Expert's answer
2022-03-23T06:43:46-0400

Horyzontalasymptotes:limxy(x)=CC2=limxx2(3ax)a+x=NohoryzontalasymptotesVerticalasymptotes:limxx0y(x)=limxx0y2(x)=+limxx0x2(3ax)a+x=+{a+x=03ax0Thusfora=0noverticalasymptotesFora0:x=averticalasymptoteHoryzontal\,\,asymptotes:\\\underset{x\rightarrow \infty}{\lim}y\left( x \right) =C\\C^2=\underset{x\rightarrow \infty}{\lim}\frac{x^2\left( 3a-x \right)}{a+x}=\infty \\No\,\,horyzontal\,\,asymptotes\\Vertical\,\,asymptotes:\\\underset{x\rightarrow x_0}{\lim}y\left( x \right) =\infty \Rightarrow \underset{x\rightarrow x_0}{\lim}y^2\left( x \right) =+\infty \Rightarrow \\\Rightarrow \underset{x\rightarrow x_0}{\lim}\frac{x^2\left( 3a-x \right)}{a+x}=+\infty \Rightarrow \left\{ \begin{array}{c} a+x=0\\ 3a-x\ne 0\\\end{array} \right. \\Thus\,\,for\,\,a=0 no\,\,vertical\,\,asymptotes\\For\,\,a\ne 0: x=-a\,\,-vertical\,\,asymptote


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