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.

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Wow thanks a lot!