Eliminating the #x# in the numerator and in the denominator gives us a hole at #x = 0#.
There is a vertical asymptote at #x= 2# and a horizontal asymptote at #y = 0#.
The intercepts are:
#y = 1/(0 - 2)#
#y = -1/2#
AND
#0 = 1/(x - 2)#
#0 = 1#; there are no x-intercepts.
#:.#There is a y-intercept at #(0, -1/2)#.
The end behaviour of the function can be found by considering some test points.
Let #x= -100# and #x = -1000#
For #x= -100#
#y = 1/(x- 2)#
#y= 1/(-100 - 2)#
#y ~=-0.0098#
For #x = -1000#
#y = 1/(-1002)#
#y = -0.000998#
As you can see, these numbers are trending towards the horizontal asymptote at #y= 0#.
We do the same near the asymptote.
Let #x = 1.75# and #x= 1.99#.
For #x = 1.75#
#y = 1/(x- 2)#
#y = 1/(1.75 - 2)#
#y= -4
For #x = 1.99#
#y = 1/(x- 2)#
#y = 1/(1.99 - 2)#
#y = -100#
So, as #x# approaches #2# from the left side, #y# approaches negative infinity. Doing the same process for the right side, you should get the as #x# approaches #2# from the right side, #y# approaches positive infinity.
We repeat the process of checking end behaviour at the right side of the function as #x# approaches positive infinity, to get that #y# tends to #y =0#.
The final graph should look like the following, with a hole at #x= 0#.
Hopefully this helps!
