Question 131510
The equation of a vertical asymptote is in the form {{{x=a}}} where {{{a}}} is a value that would make the denominator go to zero.  


To find horizontal asymptotes, compare the degree of the numerator and denominator polynomials.


If the degree of the denominator polynomial is greater than the degree of the numerator polynomial, the horizontal asymptote is {{{y=0}}}, in other words, the x-axis.  That's because, as x gets larger, the denominator is going to get larger much faster than the numerator and the entire fraction will tend to zero.


If the degrees of the numerator and denominator are the same, then, as x gets very large, the only significant difference between the numerator and denominator will be the lead coefficients.  Therefore, the horizontal asymptote is {{{y=a[n]/a[d]}}} where {{{a[n]}}} is the lead coefficient on the numerator polynomial and {{{a[d]}}} is the lead coefficient on the denominator polynomial.


If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.  There is a straight line slant or skew asymptote if the degrees differ by 1 that is found by performing polynomial long division of the denominator into the numerator.  The equation of the skew asymptote is the quotient obtained excluding any possible remainder.


The x-intercepts are given by setting the numerator equal to zero and solving.  Any zero that is also a zero of the denominator must be excluded.  This is because {{{a/b=0}}} if and only if {{{a=0}}} and {{{b<>0}}}.


y-intercepts are found by evaluating {{{f(0)}}}.


Now, for your problem:
{{{s(x)=(2x-4)/(x^2+x-2)}}}?


Vertical asymptotes: You need to find the zeros of the denominator polynomial:

{{{x^2+x-2=0}}}
{{{(x+2)(x-1)=0}}}
So the zeros are at 1 and -2, making the equations of the asymptotes {{{x=1}}} and {{{x=-2}}}


Horizontal asymptotes:  The degree of the denominator is 2 and the degree of the numerator is 1, therefore the horizontal asymptote is at {{{y=0}}}


x-intercepts:


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


And the x-intercept is at (2,0)


y=intercept:


{{{s(0)=(2(0)-4)/((0)^2+(0)-2)}}}
{{{s(0)=-4/-2=2}}}


So the y-intercept is at (0,2)


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locate(.2,1.8,YI(0,2)),
circle(0,2,.1),
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