I am giving medals. Graph f(x)= 2/(x^2-9) + 1. I'm thinking my graph should be a parabola because of the x^2, but I'm a little confused because I got three asymptotes. Any help would be much appreciated!

Question

amistre64 · Answer

an x^2 in the denominator?  this isnt going to be a parabola

amistre64 · Answer

assuming this is:

   2
------  + 1
x^2-9

the bottom zeros out at 3 and -3 giving us 2 vertical asymptotes.

anonymous · Answer

Yes I got that, so there would be 2 vertical asymptotes at -3 and 3 and a horizontal asymptote at 1? What would the graph look like?

leader · Answer

Here is your graph https://www.desmos.com/calculator/h6pvz4xflf