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

Question

Christian Antunez · Accepted Answer

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

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# A log function can only have this type of asymptote. Calculus provides the best explanation, which basically boils down to this: 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). 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{[-5.47, 26.55, -5.75, 10.27]} log{log(x+3)/log2

Sophie Adams · Answer

undefined To find the vertical asymptote for $ f(x) = \log_2(x+3) $, set the argument of the logarithm equal to zero, $ x + 3 = 0 $, and solve for $ x $. This yields $ x = -3 $. Therefore, the vertical asymptote is $ x = -3 $.

Since $ f(x) = \log_2(x+3) $ is a logarithmic function, it does not have horizontal or slant asymptotes.