Question

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

Answer 1

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]}