I'm not sure as to what you mean by "Discuss completely (as in the textbook)", but I'm assuming that you want to find the asymptotes right?
Start with the given function
Looking at the numerator , we can see that the degree is since the highest exponent of the numerator is . For the denominator , we can see that the degree is since the highest exponent of the denominator is .
Oblique Asymptote:
Since the degree of the numerator (which is ) is greater than the degree of the denominator (which is ), there is no horizontal asymptote. In this case, there's an oblique asymptote
To find the oblique asymptote, simply use polynomial division to find it. The quotient of is the equation of the oblique asymptote
___-x__________-_3___
2x^2-6x | -2x^3 + 0x^2 + 6x
-2x^3 + 6x^2
----------
-6x^2 + 6x
-6x^2 + 18x
-------------
-12x
note: in this case, we don't need to worry about the remainder
Since the quotient is , this means that the oblique asymptote is
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Vertical Asymptote:
Start with the given function
Factor out the GCF
Cancel like terms
So we're left with
where (this is where a hole occurs)
To find the vertical asymptote, just set the denominator equal to zero and solve for x
Set the denominator equal to zero
Add 3 to both sides
Combine like terms on the right side
So the vertical asymptote is
Notice if we graph , we can visually verify our answers:
Graph of with the oblique asymptote (blue line) and the vertical asymptote (green line) and the hole (the small circle drawn)