Question 1090285
IF YOU MEANT {{{f(x)=(x^3+x^2)/(x^2-4)}}} :
Factoring, we get
{{{f(x)=(x+1)x^2/((x+2)(x-2))}}}
The function is undefined for {{{x=-2}}} and {{{x=2}}} ,
and {{{highlight(x=-2)}}} and {{{highlight(x=2)}}} are the equations of the vertical asymptotes.
The value of the function is zero for {{{x=0}}} and {{{x=-1}}} ,
so {{{highlight(x=-1)}}} and {{{highlight(x=0)}}} are the x-intercepts, where {{{y=0}}} ,
and {{{y=0}}} is the y-intercept, where {{{x=0}}} .
In other words, we know the graph goes though (0,0) and (-1,0).
The numerator of the function, {{{(x+1)x^2}}} , changes sign only at {{{x=-1}}} .
The denominator of the function, {{{(x+2)(x-2)}}} , changes sign at {{{x=-2}}} and {{{x=2}}} .
All four factors are positive for {{{x>2>-1>-2}}} , nd so is {{{f(x)}}} .
As you cross {{{x=2}}} , the fucntion changes sign,
so {{{f(x)<=0}}} for {{{-1<x<2}}} , with {{{f(0)=0}}} for {{{x=0}}} .
For {{{-2<x<-1}}} , {{{f(x)>0}}} ,
and for {{{x<-2}}} , {{{f(x)<0}}} .
As for a slant asymptote, doing the division we find that
{{{(x^3+x^2)/(x^2-4)=x+1+(4x+4)(x^2-4)}}} ,
so {{{highlight(y=x+1)}}} is the slant asymptote.
This is what the graph of the function and the slant asymptote looks like:
{{{graph(300,300,-10,10,-10,10,(x^3+x^2)/(x^2-4),x+1)}}} .
With the vertical asymptotes added, it would look like this:
{{{drawing(300,300,-10,10,-10,10,
line(-2,-10,-2,10),line(2,-10,2,10),
graph(300,300,-10,10,-10,10,(x^3+x^2)/(x^2-4),x+1)
)}}} .