How do you find the vertical, horizontal or slant asymptotes for #f(x)=log_2(x+3)#?

1 Answer
Dec 12, 2016

Answer:

Vertical asymptote at #x=-3#; no other asymptotes exist.

Explanation:

Logarithmic functions will have vertical asymptotes at whatever #x#-values makes the log argument equal to 0. In this case, we will have a vertical asymptote at

#x+3=0#
#=>x="-"3#

This is the only kind of asymptote a log function can have. The best explanation comes from calculus, but essentially, it comes down to this:

  1. There can't be a horizontal asymptote because no matter how large a #y#-value you may seek, you can find an #x#-value that gives you that #y#. (If you want #log_2(x+3)="1,000,000"#, then you choose #x=2^"1,000,000"-3.#) Thus, log functions have no maximum (and no horizontal asymptote).

  2. There can't be a slant (slope) asymptote because the slopes of the tangent lines get closer to 0 as #x# goes to infinity. (The "instantaneous slope" of #log_2(x+3)# at #x# is #ln2/(x+3)#, and as #x# gets larger, this value gets closer to 0.) In other words, no matter how close to 0 you want your "instantaneous" slope to be, there will be an #x#-value that gives you that slope. Thus, log functions have no limiting rate of increase (and no slant asymptote).

So the only asymptote we have is #x="-3"#.

graph{log(x+3)/log2 [-5.47, 26.55, -5.75, 10.27]}