Can a graph of a rational function have no vertical asymptote?

Can a Graph of a Rational Function Really Have No Vertical Asymptote?

Rational functions. The name itself sounds a bit intimidating, right? But they’re really just fractions made up of polynomials, and they can do some pretty interesting things. One of the trickiest things about them is figuring out their asymptotes – those invisible lines that the graph gets closer and closer to, but never quite touches. Usually, we’re talking about vertical asymptotes, those up-and-down lines that signal a function is going haywire, shooting off towards infinity. But here’s a question that might make you scratch your head: can a rational function completely avoid having any vertical asymptotes?

Believe it or not, the answer is yes! It’s a bit like finding a unicorn, but it’s totally possible.

So, what is a vertical asymptote, anyway? Well, it happens when the bottom part of our fraction (the denominator) tries to be zero, while the top part (the numerator) is doing its own thing, staying away from zero. Think about it: you can’t divide by zero. The function just freaks out and heads for the stratosphere (or dives to the bottom of the ocean!). That’s where you get that vertical line on the graph.

But what if the denominator never equals zero? What if it’s impossible, no matter what number you plug in for x? Then, my friends, you’ve got yourself a rational function without a single vertical asymptote.

Take this function, for instance:

f(x) = x / (x² + 1)

See that denominator, x² + 1? No matter what number you square, it’s always going to be zero or positive. Add 1, and you’re always going to be greater than or equal to 1. It just can’t be zero. So, this function cruises along without any vertical asymptotes to worry about. Pretty cool, huh?

Now, here’s another sneaky way a rational function can dodge those vertical lines: removable discontinuities. Sounds fancy, but it’s actually pretty simple. Imagine you have the same thing multiplying on the top and bottom of your fraction. Like this:

f(x) = (x – 2) / (x² – 4)

Aha! We can factor that denominator:

f(x) = (x – 2) / ((x – 2)(x + 2))

Notice anything? We’ve got (x – 2) on both the top and the bottom. Now, technically, the function is undefined when x = 2, because that would make the denominator zero. But, if we cancel out those (x – 2) terms (and we’re careful to remember that x can’t really be 2), we get:

f(x) = 1 / (x + 2), x ≠ 2

What we’ve done is created a “hole” in the graph at x = 2. It’s like the function skips over that point. The only vertical asymptote we’re left with is at x = -2, where the simplified denominator is zero.

Of course, there’s a catch. If you have something like (x – a) on the bottom more times than you have it on the top, you’re still going to get an asymptote at x = a. It’s all about who has the most “power.”

So, there you have it. Rational functions can exist without vertical asymptotes, either because their denominators are always positive, or because they have sneaky “holes” in their graphs. Math is full of surprises, isn’t it?