Question 1090285
{{{f(x)=(x^3+x^2)/(x^2-4)}}} -> form {{{f(x) = p(x) / q(x)}}}

you are given a rational function written as the ratio of two polynomials where the denominator {{{q(x)}}} isn't zero

Vertical Asymptotes:
An asymptote is a line that the curve approaches but does not cross. The equations of the vertical asymptotes can be found by finding the roots of {{{q(x)}}}. Completely ignore the numerator when looking for vertical asymptotes, {{{only}}} the {{{denominator}}}{{{ matters}}}. 

{{{x^2-4=0}}}

{{{x^2=4}}}

{{{x=sqrt(4)}}}

{{{x= 2}}} or {{{x=-2}}}

so, vertical asymptotes: {{{x= 2}}} and {{{x=-2}}}

Horizontal Asymptotes:

The location of the horizontal asymptote is determined by looking at the degrees of the numerator ({{{n}}}) and denominator ({{{m}}}).

    If {{{n<m}}}, the x-axis, {{{y=0}}} is the {{{horizontal}}} asymptote.
    If {{{n=m}}}, then {{{y=an / bm}}} is the {{{horizontal}}} asymptote. That is, the ratio of the leading coefficients.
    If {{{n>m}}}, there is {{{no}}} horizontal asymptote. 
However, if {{{n=m+1}}}, there is an {{{oblique}}} or {{{slant}}} asymptote.

in your case,{{{n=3}}} and {{{m=2}}}; so, {{{n>m}}} which means there is {{{no}}} horizontal asymptote

Oblique Asymptotes:

When the degree of the numerator is exactly one more than the degree of the denominator, the graph of the rational function will have an oblique (slant) asymptote. 

in your case {{{n=3}}} and {{{m=2}}} and {{{n=m+1}}}; so, you have a slant asymptote
To find the equation of the oblique asymptote, perform long division (synthetic if it will work) by dividing the denominator into the numerator.

{{{f(x)=(x^3+x^2)/(x^2-4)}}}

------------{{{highlight(x+1)}}}
{{{ (x^2-4)}}}/{{{x^3+x^2}}}
-------------{{{x^3-4x}}}
----------------{{{0+x^2}}}
-----------------{{{x^2-4}}}
---------------------{{{4}}}

so, {{{f(x)=(x^3+x^2)/(x^2-4)}}} is asymptotic to {{{f(x)=highlight(x + 1)}}}


{{{drawing(600, 600, -10, 10, -10, 10,
green(line(2,10,2,-10)),green(line(-2,10,-2,-10)),
 graph(600, 600, -10, 10, -10, 10, (x^3+x^2)/(x^2-4),x+1)) }}}