Topic 1.7: Rational Functions and End Behavior

A visitor from another star

On the first of July, 2025, a telescope in Chile caught a faint smudge creeping across the stars, and astronomers soon realized it was something we’d only ever seen twice before. It came from another star. It’s called three-I ATLAS, and it was falling toward the Sun at more than sixty kilometers every second, far too fast for the Sun ever to hold onto it.

For a few months people even wondered whether it was an alien spacecraft, and NASA’s answer was that it’s a comet, just one from somewhere else. Out there between the stars, it travels in a dead-straight line.

Then, at the end of October, it swung around the Sun, the heat lit it up with two glowing tails, and the Sun’s gravity bent its path. Now it’s leaving on a brand-new straight line, out of our solar system, and in NASA’s words, never to return. Its path has a formula, and that formula is a fraction, one polynomial divided by another, and hiding inside it is the exact line this visitor will ride out of the solar system, forever.

By the end of this lesson, you’ll be able to take the formula for this visitor’s path and find that line yourself. And you’ll do it from just two of its terms. To get there, we need to know one thing. What is a fraction like that really measuring?

Can you climb back?

So let’s start somewhere closer to home. Your win rate. Say you’ve played a hundred ranked games and you’re forty and sixty, so you’ve won forty percent of the time. But you’ve been practicing, and from now on you win three of every four games. So how many more games until your win rate climbs all the way to seventy-five percent?

If you play g more games and win three quarters of them, your wins are forty plus point seven five g, your games are a hundred plus g, and your win rate is wins divided by games. A hundred more games puts you at fifty-seven and a half percent, a thousand gets you near seventy-two, and ten thousand gets you to seventy-four point six five. Even after a hundred thousand games, more than almost anyone will ever play, you’re at seventy-four point nine seven, still creeping up, and you never actually reach seventy-five.

So look at the fraction after ten thousand games, which is forty plus seven and a half thousand, over a hundred plus ten thousand, and the forty and the hundred hardly matter anymore. So far out it acts like point seven five g over g, which is just point seven five. Your past is the constant term, and in the long run, constants lose.

Why never? A true seventy-five percent player would have seventy-five wins by now. You have forty. Winning at their pace never wins back those thirty-five. It only spreads them over more games.

Now you try one. Your friend has won thirty games and lost seventy, and from now on she wins four of every five, so what win rate is she heading toward, as a percent?

Yours and hers. They are the same kind of fraction. A win rate like that is a rational function, which is the name for one polynomial divided by another. Every rational function compares how big its top is with how big its bottom is, and I like to picture that as a question, how many copies of the bottom fit inside the top?

Take six x squared plus five, all over two x squared plus one. At x equals one the top is eleven and the bottom is three, and three fits into eleven about three and two thirds times.

Now let x grow. At ten, about three point zero one copies fit, and at a thousand the top is six million and five and the bottom is two million and one, and three copies fit almost exactly.

You saw last lesson that far out a polynomial acts like its leading term, and here both of them do, so the count settles on six x squared over two x squared, which is three. Far enough out, only the leading terms are left. Both of these crept toward a number they never reach, and there’s a famous puzzle about exactly that, one that fooled Albert Einstein.

The puzzle that fooled Einstein

About ninety years ago the psychologist Max Wertheimer sent Albert Einstein a puzzle about an old, clattery car that has to drive two miles, one mile up a hill and one mile back down. The car is so old it can only climb at fifteen miles per hour, and the question is how fast it has to come down if it wants to average thirty for the whole trip.

So, what do you think? The car comes down the hill. How fast does it have to go? To average thirty miles an hour over two miles, the whole trip has to fit in four minutes, because thirty miles an hour is half a mile every minute. The light on the road burns away at that pace, so it reaches the top in two minutes.

Now watch the clock while the car climbs at fifteen, which is only a quarter mile a minute. At two minutes the light is already over the top, and the car is only halfway up. It reaches the top at exactly four minutes, just as the light runs out. The climb used up the entire budget, so there’s no time left for the way down, and Einstein wrote back that he didn’t notice that until he actually did the calculation.

Come down at forty-five and the trip averages twenty-two and a half, at ninety it’s about twenty-five point seven, and even at six hundred, as fast as a jet airliner, it’s only about twenty-nine point three. Break the sound barrier and it’s still not thirty, and even at the speed of light the average is twenty-nine point nine nine nine nine nine nine three, so thirty is a ceiling this trip can never reach.

You can’t reach thirty, but you can get as close as you like, so try this. Push the descent speed until the trip averages within a tenth of a mile per hour of thirty.

Let’s put this car on paper, so we can see exactly why thirty is the number it’s chasing. Time is distance over speed, so the trip takes one fifteenth of an hour going up plus one over v coming down. Two miles divided by that total time is the average speed. To clear the little fractions, we multiply the top and the bottom by fifteen v.

On top, two times fifteen v is thirty v. On the bottom, fifteen v over fifteen is just v, and fifteen v over v is fifteen. So it becomes thirty v over v plus fifteen. And look at that, it’s another rational function, degree one over degree one, the same kind of tie as the win rate.

For huge speeds the fifteen is nothing next to v, so the leading terms take over, thirty v over v, which is thirty, the ceiling we kept running into. On a graph the average climbs fast at first and then flattens out under a line at thirty, which we call a horizontal asymptote. We write it as a limit, the limit of A of v as v goes to infinity equals thirty, and you read that arrow as “v increases without bound.”

What that limit promises is exactly what you did with the throttle. Pick any closeness you like, a tenth, a millionth, and there’s a speed past which the average gets that close to thirty and stays that close for good. This is exactly how the exam asks it. They give you the limit and ask what it means, and a full-credit answer says that as the descent speed increases without bound, the average speed for the trip approaches thirty miles per hour.

Top versus bottom

The win rate and the hill were both ties, where the top and the bottom have the same degree, so the leading terms cancel down to a plain number. But the top and the bottom don’t have to be evenly matched. Let’s turn the dial down so the top is six x plus five. Now the bottom grows like x squared and the top only like x, and the count gets crushed, point three, point zero three, point zero zero three, on its way to zero.

So when the bottom has the higher degree, the bottom wins. And the whole function heads to zero, so its horizontal asymptote is y equals zero. Now push the top up to six x cubed plus five. The copies pile up, thirty, three hundred, three thousand, and the graph never levels off. It runs along the line y equals three x, which is six x cubed over two x squared.

One more power on top and the leading terms give three x squared, so both ends shoot up like a parabola. When the top wins, the function behaves like the polynomial from its leading terms.

Your turn. Here’s the dial again, still at degree four. Set the top’s degree and its leading coefficient so the graph levels off at y equals negative three. So the two degrees decide how the contest ends. On the AP exam you’ll see the winner called the polynomial that dominates.

Now the whole contest in symbols. Call the top’s leading term a x to the m, and the bottom’s b x to the n. Here, a and b are the leading coefficients, and m and n are the degrees.

Far out, the function acts like a x to the m over b x to the n. That simplifies to a over b, times x to the m minus n. If the bottom’s degree is bigger, that power is negative and the whole thing goes to zero. If the degrees match, the x’s cancel and you’re left with a over b.

And if the top’s degree is bigger, you’re left with a polynomial, and when the top wins by exactly one, that polynomial is a line, and the graph rides a slant asymptote with that same slope. And this works at both ends, since x heading to negative infinity is just as huge in size, so a tie has the same limit, a over b, in both directions.

Here’s one that’s written to trip you up, four plus three x minus two x squared, over five x squared plus one. The leading term isn’t the one written first, it’s the one with the highest power, and on top that’s negative two x squared, hiding at the very end. So the leading terms are negative two x squared over five x squared, which is negative two fifths, and that’s the horizontal asymptote at both ends.

Factored forms hide their leading terms too, so there you multiply the leading piece of each factor. For example, take x minus three, times four x plus one, over two x plus five, times x minus one.

On top, the leading pieces are x and four x. Together they make four x squared. On the bottom, two x times x makes two x squared. Four x squared over two x squared is two, so the asymptote is y equals two. Here’s one for you, and take your time with it. What’s the horizontal asymptote of this one?

The way out

Let’s go back to our visitor. As it leaves, the Sun keeps pulling on it, with a strength of a number divided by r squared, where r is its distance from the Sun. That’s a constant over a growing bottom, so the bottom wins and the pull fades toward zero, and out at Voyager 1’s distance it’s about thirty thousand times weaker than at the Earth’s distance from the Sun.

The Sun never lets go, but its grip heads to zero. Now I’m going to turn the picture, so the visitor came in straight down from the top. Then I’ll lay a pair of axes over space, measured in astronomical units. One unit is the distance from the Earth to the Sun.

In this picture, its real path, worked out from NASA’s measurements of its orbit, is y equals point two one minus two point nine five x squared, all over x. The top has degree two and the bottom has degree one, so the top wins by exactly one, and that means the path settles along a line.

Now you get to be the astronomer. Use just those two leading terms and lay down the line this visitor will ride out of our solar system.

Now watch the gap between the path and the line. It’s about point two at x equals one, about point zero four at five, and less than a hundredth by thirty. That gap is the leftover piece of the formula, point two one over x, and it’s a constant over a growing bottom, so it dies. It’s the Sun’s bend fading away, and what’s left forever is the leading terms.

So the visitor’s whole path is a line plus a leftover, and only the line survives. If we split the fraction, the path is point two one over x, minus two point nine five x, a leftover that dies plus the line it rides.

For our visitor, the leading terms gave the exit line exactly. Its top had no next term to shift the line. This one does: x cubed plus two x squared plus one, all over x squared plus one.

The leading terms are x cubed over x squared, which is x, so you’d expect the graph to hug the line y equals x. But the graph sits about one point eight nine above that line at ten, one point nine nine above it at a hundred, and one point nine nine nine above it at a thousand. It’s riding a line two units higher, y equals x plus two. It has the same slope as y equals x, so the two lines are parallel.

The next term, two x squared, lifts the line by two. Lesson one point eleven finds that shift for any bottom, with long division. The ends still run off to infinity: up on the right, down on the left. A slant asymptote never makes a function level off. It only tells you which way the ends are heading.

One more thing to watch when the top wins: the signs. Take negative x cubed plus four, over x squared plus three. Its leading terms give negative x cubed over x squared, which is negative x. Far to the right, negative x is hugely negative, so the graph falls. Far to the left, negative x is hugely positive, so the graph rises.

An odd power flips its sign between the two ends. An even power, like x squared, never does. Here’s one where the top wins by a lot. Which of these describes its end behavior?

Can it touch?

Our visitor never touches its exit line, and Einstein’s car never reaches thirty, so it’s really tempting to decide that a graph can never cross its asymptote. Lots of explanations online say exactly that, and for a horizontal asymptote it’s wrong. Take two x squared plus eight x, over x squared plus four. It’s a tie, two x squared over x squared, so the horizontal asymptote is y equals two.

But does the graph ever actually equal two? Set it equal to two and multiply both sides by x squared plus four, and the two x squareds cancel, which leaves eight x equals eight, so x equals one.

So it crosses its own asymptote right there at x equals one, climbs to about three and a quarter, and then turns around and comes back down toward two. And from about x equals seventy-nine on, it stays within a tenth of two and never leaves, so y equals two is still its horizontal asymptote.

A horizontal asymptote is a promise about the ends of the graph. It is not a wall in the middle. So in the middle, the graph can cross it as often as it likes.

Only the leading terms are left

So here’s the whole lesson in three pictures. When the bottom has the higher degree, the bottom wins, and the function heads to zero. When it’s a tie, it levels off at the ratio of the leading coefficients. That’s what your win rate and Einstein’s car did.

And when the top wins, it acts like the polynomial from the leading terms, riding a slanted line if it wins by exactly one. Find the leading terms by their power and not their position, and remember that a horizontal or slant asymptote is about the ends, so the middle can cross it.

And that brings us back to our visitor. You took its formula, kept just two terms, and found the line it’s riding out of our solar system. In November 2026 Voyager 1, the farthest spacecraft we’ve ever launched, crosses one light-day from Earth, and our visitor is heading out more than three times faster, along a line you can now write down. Far enough out, only the leading terms are left. And now you know how to find them.