Question 734948
to find the slant asymptotes of

a. {{{(x^3+x^2)/(x^2-4)}}}

{{{x^2(x+1)/(x^2-2^2)}}}

{{{x^2(x+1)/((x-2)(x+2))}}}

The vertical asymptotes (and any restrictions on the domain) come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve.

{{{((x-2)(x+2))=0}}}

if {{{x-2=0}}}...=>...{{{x=2}}}
if {{{x+2=0}}}...=>...{{{x=-2}}}

Then the domain is all {{{x-values}}} other than ± {{{2}}}, and the two vertical asymptotes are at
{{{x = 2}}} and {{{x=-2}}}.

Since the degree is greater in the denominator than in the numerator, the {{{y-values}}} will be dragged down to the {{{x-axis}}}, and the horizontal asymptote is therefore "{{{y = 0}}}"  (the {{{x-axis}}}) .

The slant asymptote:  use the long division to find it

----------{{{x+1}}}
--------|{{{x^3+x^2}}}
{{{x^2-4}}})..{{{x^3-4x}}}
_________________
----------{{{0+4x+x^2}}}
------------ {{{x^2-4}}}
________________
---------------{{{4x+4}}}

---------------{{{4(x+1)}}}

so, since {{{x+1}}} is a factor of the numerator

the slant asymptote:  {{{y = x + 1}}}


{{{drawing(600,600, -10,10, -10, 10, blue(line(2,10,2,-10)),  blue(line(-2,10,-2,-10)), graph( 600,600, -10,10, -10, 10, (x^3+x^2)/(x^2-4),x+1),0)) }}} 


b. {{{(2x^2+10x-12)/(x^2+x-6)}}}

horizontal asymptote: -> to {{{-12/-6=2}}} as {{{x}}}-> {{{infinity}}} or {{{-infinity}}}

so,horizontal asymptote is {{{y=2}}}

continue factoring

{{{(2x^2+10x-12)/(x^2+x-6) }}}

{{{2(x^2+5x-6)/(x^2+x-6) }}}

{{{2(x^2-x+6x-6)/(x^2+3x-2x-6) }}}

{{{2((x^2-x)+(6x-6))/((x^2+3x)-(2x+6)) }}}

{{{2(x(x-1)+6(x-1))/(x(x+3)-2(x+3)) }}}

{{{2((x+6)(x-1))/((x-2)(x+3)) }}}

find values that make denominator equal to zero:

if{{{x-2=0 }}}....->...{{{x=2}}}

if{{{x+3=0 }}}....->...{{{x=-3}}}

these are vertical asymptotes

-there is no slant asymptote in this case


{{{drawing(600,600, -10,10, -10, 10, blue(line(2,10,2,-10)),  blue(line(-3,10,-3,-10)), graph( 600,600, -10,10, -10, 10, (2x^2+10x-12)/(x^2+x-6),2)) }}}