Rational Function Graphing Question

how do you determine were a graph intersects in the horizontal, oblique, slant asymtote.

an example how would you decide that for a problem like this

g(x)= x^2/x^2+x-6

g(x) = x²/(x²+x-6) = x²/((x+3)(x-2))

To find vertical asymptotes, find where denominator = 0 ----> when x = -3 and x = 2
In all cases numerator of g(x) >= 0
As x → -3 from the left, denominator → 0 from positive side, so g(x)→∞
As x → -3 from the right, denominator → 0 from negative side, so g(x)→-∞
As x → 2 from the left, denominator → 0 from negative side, so g(x)→-∞
As x → 2 from the right, denominator → 0 from positive side, so g(x)→∞

To find horizontal asymptote, find limit as x approaches ∞ and -∞
lim[x→∞] x²/(x²+x-6) = lim[x→∞] 1/(1 + 1/x + 6/x²) = 1/(1+0+0) = 1
lim[x→-∞] x²/(x²+x-6) = lim[x→-∞] 1/(1 + 1/x + 6/x²) = 1/(1+0+0) = 1

There is no oblique (or slant) asymptote.

To find where graph crosses horizontal asymptote (y=1), simply solve
g(x) = 1
x²/(x²+x-6) = 1
x² = x² + x - 6
0 = x - 6
x = 6

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On interval (-∞, -3), g(x) goes from just above 1 to +∞
http://i54.tinypic.com/dqouvr.png

On interval (2, ∞), g(x) goes from +∞ to just below 1, crossing over y=1 at x=6
http://i53.tinypic.com/3wkk5.png

On interval (-3, 2), g(x) = x²/((x+3)(x-2)) has numerator >= 0 and denominator < 0
So g(x) <= 0, and g(x) = 0 when x = 0
On this interval, g(x) goes from -∞ up to 0 (at x = 0), then to -∞
http://i51.tinypic.com/23hnhgi.png

I presume that it is g(x) = (x^2) / (x^2 + x - 6)

The y intercept is found by letting x = 0 and this of course produces y = 0.

There is no other x intercept because y = 0 has only x = 0 as solution.

This function has a horizontal asymptote because the power of the numerator is the same as the power of the denominator. (It would be a slant asymptote if the power on top was one higher and no asymptote if two or more higher.)