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

Dec 12, 2016

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

$x + 3 = 0$
$\implies x = \text{-} 3$

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 ${\log}_{2} \left(x + 3\right) = \text{1,000,000}$, then you choose $x = {2}^{\text{1,000,000}} - 3.$) 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 ${\log}_{2} \left(x + 3\right)$ at $x$ is $\ln \frac{2}{x + 3}$, 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 $x = \text{-3}$.

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