Firstly I would rewrite it as (x^2-9)/(x^2-9) + 9/(x^2-9)
and then factorise
to get y = 1+ 9/(x+3)(x-3)
We can tell a lot about this function purely from its domain
1. Domain
The domain is all real x, except x is not equal to 3 or -3 for this would make the denominator 0.
2. Odd/Even Function
Substituing f(-x) gives the same function value as f(x), so hence the graph is symmetrical about the y-axis.
3. Asymptotoes
Taking the limit to infinity gives y=1 resulting in a horizontal asymptote at y=1. Now we know also the vertical asymptotes are at x=3 and x=-3, see Drongoski's post ^
Justification of why Drongoski's method works
Another LONGER way to find them is to express x in terms of y
in this case we get (x-3)(x+3) = 1/(y-1)
Letting y approach infinity (postive or negative makes no real difference here)
gives (x-3)(x+3) = 0 giving the asymtote values.
4. Range
The range is y is not equal to 1. For this would mean that 9/(x+3)(x-3) would be equal to 0 which is impossible.
Also note that the range 0<y<=1 is also not a valid range either, since that f(0) = 0 and the function is even, and noting that f(2) < 0 and f(4) > 1.
Now we can graph since we have enough information. Note this approach does not involve calculus.
The graph starts asymptotic to the line y=1, it then increases as x approaches -3 to positive infinity asymptotically.
The graph "resumes" asymptotic to the line x=-3 approaching from negative infinity. It then reaches the local maximum of (0, 0)
and the rest of the graph is a reflection about the y-axis.
There is a free graphing program called Geogebra. You can also use it on the web, its fairly useful.
The things you need to determine to sketch a graph are:
Domain/Range
Odd/Even Function (i.e Symmetry)
Asymptotes
and then if you still can't graph, then use Calculus.
If the questions asks for turning points, then calculus is required.
DDDD