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For each functon find the following (if they exist).

     a. End Behavior including the equations of horizontal or slant asymptotes.

     b. Vertical Asymptote(s). Distinguish between VA and Holes.

 

\(f(x)=\frac{-3x^3}{x^3-4x}\)

AdamTaurus  Nov 15, 2017
 #1
avatar+94117 
+1

For each functon find the following (if they exist).

     a. End Behavior including the equations of horizontal or slant asymptotes.

     b. Vertical Asymptote(s). Distinguish between VA and Holes.

 

 

\(y=\dfrac{-3x^3}{x^3-4x}\\ y=\dfrac{-3x^3}{x(x-2)(x+2)}\\~\\ so\;\;x\ne\pm2,\;\;\;x\ne0\\~\\ y=\dfrac{-3x^3}{x^3-4x}\\ \)

 

 

 

\(y=\dfrac{-3x^3\div x^3}{(x^3-4x)\div x^3}\\ y=\dfrac{-3}{1-\frac{4}{x^2}}\\ \displaystyle\lim_{x\rightarrow 0^\pm}y=\frac{-3}{1-\infty}=0 \quad \text{Hole at (0,0)} \\\displaystyle\lim_{x\rightarrow \pm\infty}y=\frac{-3}{1-0}=-3\\~\\ \text{Let }\delta\;\;\text{ be a miniscule positive number.}\\ \displaystyle\lim_{x\rightarrow2^+}y=\frac{-3}{1-(1-\delta)}=\frac{-3}{\delta}=-\infty\\ \displaystyle\lim_{x\rightarrow-2^-}y=\frac{-3}{1-(1-\delta)}=\frac{-3}{\delta}=-\infty\\ \displaystyle\lim_{x\rightarrow2^-}y=\frac{-3}{1-(1+\delta)}=\frac{-3}{-\delta}=\infty\\ \displaystyle\lim_{x\rightarrow-2^+}y=\frac{-3}{1-(1+\delta)}=\frac{-3}{-\delta}=\infty\\ \)
 

Hole at (0,0)

Vertical asymptotes at  x=+2 and x=-2

Horizontal asymptotes at  y=-3

 

Here is the graph

 

Melody  Nov 15, 2017

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