# Thread: A horizontal asymptote I'm not seeing...

1. ## A horizontal asymptote I'm not seeing...

Alright, in the back of the book it says the horizontal asymptote is $y = 2$. I know that's true because when I went and graphed it, the lines were scared of that number.

Anyway, here is the equation:

$f(x) = 2+ \frac{1}{x}$

$n < m$ makes $y= 0$ is what I thought but there's a two.

No idea what to do here.

Wouldn't getting rid of the bottom make it:
$f(x) = 2+ \frac{1}{x}$
$f(x) = 2x+1$

Unless...

$\frac{2}{1}+\frac{1}{x}$

$\frac{2x}{x}+\frac{1}{x}$

$= \frac{2x+1}{x}$

which makes the horizontal asymptote 2...

Never saw an example period of that though.

2. I don't understand how you're "getting rid of the bottom"...? This isn't an equation that you're solving, so you can multiply through by x or something; it's a functional statement containing a fraction.

To find the horizontal asymptote, restate the rational function as one combined fraction:

$2\, +\, \frac{1}{x}\, =\, \frac{2x}{x}\, +\, \frac{1}{x}\, =\, \frac{2x\, +\, 1}{x}$

Then apply the asymptote rules they've given you.

3. $y=f(x)$
$f(x)=\frac{1}{x}$

picture the graph of this in your mind. you know it has two asymptotes x=0 and y=0.

$f(x)+2$ is a vertical translation so move all points up. the Y axis is still an asymptote but the x axis has moved up 2.

$\begin{array}
\\\lim
\\x\rightarrow+\infty
\end{array}
$