Where does the horizontal asymptote lie for f(x) =
5 − 2x/2 − x?
y = 1
y = 2
y = 3
y = 5

Question

Where does the horizontal asymptote lie for f(x) =
5 − 2x/2 − x?
y = 1
y = 2
y = 3
y = 5

zepdrix · Answer

$$\Large\rm f(x)=\frac{-2x+5}{-x+2}$$We use limits to find out what's happening on the `ends` of a function.
This tells us about horizontal asymptotes.

The idea is basically, let's plug in a huge value for x and see what's happening way to the right side of the graph.

If we plug in x=12bajillion we get,$$\Large\rm \approx\frac{-2(12bajillion)}{-(12bajillion)}$$The +5 and +2 are sooooo insignificant at this point that we won't even worry about them.

zepdrix · Answer

If you cancel some stuff out, you'll see that we're approaching the value $\Large\rm 2$, yes?

zepdrix · Answer

So the shortcut is ... when the leading terms (highest power of x) are equal, the horizontal asymptote will be the `ratio` of their coefficients.$$\Large\rm \lim_{x\to\infty}\frac{-2x+5}{-x+2}=\frac{-2}{-1}$$