Many students have difficulty with the graph transformation of oblique asymptote. Consider the oblique asymptote y = x-1 (red line)
i) y= 1/ f(x)
f(x) approaches infinity as x approaches infinity. 1 divided by infinity is 0.
Hence oblique asymptote y=x-1 becomes horizontal asymptote y= 0
For y= 1/ f(x), any oblique asymptote y=ax+b in f(x) will become horizontal asymptote y= 0
ii) y= f ‘ (x)
The gradient of y= x-1 is 1. Hence oblique asymptote y=x-1 becomes horizontal asymptote y= 1
For y= f ‘ (x), any oblique asymptote y=ax+b in f(x) will become horizontal asymptote y= a
iii) y= f(2x+1)
Let original asymptote be f(x) = x-1
Therefore, f(2x+1) = (2x+1)-1 = 2x
Hence oblique asymptote y = x-1 is transformed to y=2x.
This method can be applied to any oblique asymptote.
Wow thanks a lot!