We’ve been studying graphs of rational functions in Precalculus.

Me: “Take 1 minute with your group: what will the graph of y = x/(x+1) look like?”

One group, during discussion, asserted that it had to be a line, using a sort of process of elimination: it’s not a parabola, it’s not cubic, it’s not a hyperbola.

Interesting, right? Why does this seem like a linear equation? I guess that it sort of looks like one…

## 8 replies on ““y=x/(x+1) has got to be a line.””

Well, except for the discontinuity stuff happening around x=-1 ,this is after all essentially the relatively featureless graph of y = 1, right? In the big picture way, considering all the numbers from -infinite to infinite (or if that’s kind of difficult, over the open interval that ends with plus/minus Avogadro’s number or plus/minus a google) that we could possibly substitute for x.

Looked at that way, “linear” is a fairly good approximation in the grand scheme of things.

Wouldn’t that be something like y = (x+1)/(x+1)?

When I said “around” x=-1, rather than “at” x =-1, it was deliberate.

I literally meant the neighborhood around x=1. The size of the neighborhood depends on how many decimal points you require the actual value to be calculated to before you’ll accept the fact that it rounds off to 1.

No, Michael’s saying that the function y = (x+1)/(x+1) gives the line you’re describing, but not y = x/(x+1). The discontinuity is for

preciselyx = -1, because that is the only point at which we’re dividing by zero. I think you may want to check the math in your sample points again. 😉Well, the x terms are both to the 1st power, so…linear! 😀

But seriously, it’s not a line. While lim (x –> infinity) of x/(x+1) = 1, that’s not the same thing as y = 1. Here are some points from y = x/(x+1)

(0, 0)

(1, 0.5)

(2, 0.66…)

(3, 0.75)

(4, 0.8)

These points clearly do not make a straight line, so we can reasonably assume that the student didn’t plug anything in and just went “first power = line” after eliminating quadratics and cubics. (I’m guessing someone also needs a little talk on the existence of hyperbolas that are rational functions!)

Yes, and someone else might benefit from a little talk on the benefits of making approximations to judge whether or not something matters a whole lot or can, for practical intents and purposes, be largely ignored in a particular situation.

I’m well aware that the graph of x/(x+1) is actually a hyperbola, but you don’t have to plug in very large values in x before the differences in y become rather boring to plot on a graph, for instance. For practical purposes, y = 1 is a good enough approximation to the function if the absolute value of x is big enough.

I’m looking at this debate from an engineering background rather than a mathematical one.

Can you think of any practical application in which the difference between the linear approximation y=1 and the actual hyperbolic path matters, or do you just like to quibble over technicalities?

Have you tried evaluating the function at x=Avogadro’s number? Do you see my point?

But what is the situation we’re looking at here? Is it one in which x = Avogadro’s number is feasible? (Probably not, as that would imply we’re keeping track of individual atoms, which is highly impractical.) Is it one in which x = a million? A thousand? Or maybe x = 0.3?

Engineers do have to strike a balance between precision and practicality, yes. But you still have to have a halfway-decent idea of

whatyou’re approximating before you do it, don’t you?Agreed, and that’s why it is important that the class being discussed in the original post learn that the fumction is indeed hyperbolic, isn’t it? However, the asymptotic behavior of hyperbolas might account for the conclusion the original students reached, especially if they happened to only test this function using valurs that were too large to make its hyperbolic features interesting, don’t you agree?