Rational Functions

Question

y = (x^2+7x-4)/(x-2)

Since the degrees of both the numerator and denominator are the same, I would use the leading coefficient, and both the rational functions start with 1, so would the horizontal asymptote be 1?

Also don't get how to find the horizontal asymptote of the following :

y = (2+(x+3)/(x-9)*(5-(x^2+6x-3)/(x^2-8x)

y = ((x+5)/(x^2+6x-9))*((3x^2+2x+8)\((x^2+7x-9))

Answer 1

How can you say "the degrees of both the numerator and denominator are the same", when the numerator clearly shows an x^2 term and the denominator only an x term ?
A division of x^2 + 7x - 4 by (x-2)
= x+9 + 14/(x-2)

so as x ---->large we converge on y = x+9
so there is no horizontal asymptote, but rather an oblique one of y = x+9
There is a vertical asymptote of x = 2

for y = (2+(x+3)/(x-9)*(5-(x^2+6x-3)/(x^2-8x)

both (x+3)/x-9) and (x^2 + 6x - 3)/(x^2-8) approach a value of 1 for each as x becomes large
so as x --->∞ , you are left with
y = (2 + 1)*(5-1)
= 12
so we have a horizonatal asymptote of y = 12

the last one:
as x --->∞, I see it as
y = (1/x)*(3/1) = 3/x
which zero infinity as x --->∞

So the horizontal asymptote is y = 0

Answer 2

sorry that was the wrong question I meant the degrees of these one were both the same:

y = x^2+5x-6/x-3x^2+9

the leading coefficient is 1, so the vertical asymptote would be one?

Answer 3

for the last two you did how do you know the x is becoming large like that whole infinity thing?