Question 1027061
I assume you meant
{{{f(x)=(x^3+2x^2-4x-8)/(2x^2-8)}}} .
The numerator and denominator of that rational function can be factored.
{{{f(x)=(x^3+2x^2-4x-8)/(2x^2-8)=((x^2-4)(x+2))/2(x^2-4)}}} .
When {{{x^2-4=0}}}<--->{{{system{x=-2,"or",x=2)}}} the  denominator is zero,
and the function has no defined value.
 
a) So, the values {{{system{x=-2,"and",x=2)}}} are excluded from the domain of the function.
The function can be simplified as
{{{f(x)=(x+2)/2=(1/2)x+1}}} for {{{system{x<>-2,"and",x<>2)}}}  .
The graph of {{{f(x)}}} is a straight line with two holes,
and the behavior of {{{f(x)}}} near the excluded values (and everywhere else in its domain)
is just like the behavior of {{{g(x)=(x+2)/2=(1/2)x+1}}} .
In other words, {{{lim(x->-2,f(x))=(-2+2)/2=0}}} and  {{{lim(x->2,f(x))=(2+2)/2=2}}} .
 
b) The y-intercept of {{{f(x)}}} is {{{f(0)=(0+2)/2=2/2=1}}} ,
or the point (0,1) .
{{{f(x)}}} is never zero, because where the function exists
it is equal to {{{g(x)=(x+2)/2}}} ,
and {{{g(x)=0}}} only for {{{x=-2}}} which is not part of the domain of {{{f(x)}}} .
So, there is no x-intercept. The graph of function does not intersect the x-axis; it jumps over it.
 
c) Since everywhere in its domain {{{f(x)=(1/2)x+1}}} ,
the line {{{y=(1/2)x+1}}} could be considered the slant asymptote. There is no other asymptote.