Thread: Horizontal asymptote

Hi,
I have just started precal and I can't understand well how to find the horizontal asymptote for f(x)=(4|x-2|)/x+1. Can somebody help me. Thank you.

Originally Posted by cuteisa89

Hi,
I have just started precal and I can't understand well how to find the horizontal asymptote for f(x)=(4|x-2|)/x+1. Can somebody help me. Thank you.

You need to think about what happens when x goes to + infty and also
to -infty.

For large x:

f(x) = (4x-8)/(x+1)

divide top and bottom by x:

f(x) = (4-8/x)/(1+1/x)

so as x -> infty, f(x)->4 (as 8/x and 1/x both go to zero). So f(x)=4 (or y=4)
is an asymtot for large x.

For large negative x, and large |x|:

f(x) = -(4x-8)/(x+1)

divide top and bottom by x:

f(x) = (-4+8/x)/(1+1/x)

so as x -> infty, f(x)->-4 (as 8/x and 1/x both go to zero). So f(x)=-4 (or y=4)
is an asymtot as x->-infty.