Question 977183
f(x)= x^2-3x-18/(x^2-4) 
the denominator cannot be zero, so x^2-4 cannot be 0, and x cannot be 2 or -2.

y-intercept is when x=0, and that is -18/-4 or 4.5
x-interceps found by factoring   (x-3)(x+6)/(x-2)(x+2); x= -3 and x=+6 will make the numerator 0 and therefore make y=0.
vertical asymptote is where x= 2 or -2.  At that point, the function goes to either + or - infinity.

horizontal asymptote is where as x becomes infinite, the function approaches a certain value.  The squares are what drive the function at infinity.  x^2 is so much larger than x, that you can ignore anything that isn't a square.  The function then becomes x^2/x^2 or 1.   y=1 is the horizontal asymptote. 

Graphing by hand, I would draw in the asymptotes, the y-intercept is -4.5, the roots are 2 and -2, and you look at values less than -2, between -2 and 2, and greater than -2.

Greater or less than minus 2, and the function approaches the line y=1.
Less than -2, say -1, f(x)=-14/-3 or +4 2/3.
Less than +2, say +1, f(x)=-20/-3 or +6 2/3

At minus 2.0001, for example, the numerator is about  4-6-18=-20.  The denominator is 4.0004-4 or 0.0001.  Large negative numerator, small positive denominator.  The function has a large negative value.  At -1.9999,  the denominator is small negative, so the value becomes very large.  The same occurs at +1.9999 and +2.0001.  You know near an asymptote that the function will go to infinity.  What you do is look to see what the sign of both the numerator and denominator are.  You don't really need to calculate them, just know that the denominator will become very close to zero.  The question then becomes, close to from negative side? or close to from positive side?


{{{graph(300,300,-10,10,-10,10,(x^2-3x-18)/(x^2-4))}}}