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We've seen some huge, lumbering polynomials before. Dinosaur-sized polynomials, even, but less extinct. So the idea of putting one polynomial on top of another, to make a rational function, seems terrifying. It's a good thing we know that there's no reason to worry, though; we've messed with rational functions before.
Now we're going to unlock their secrets. No matter how complicated a rational function might appear, it will become very tame and predictable once we figure out where it asymptotes and zeros are. And that is the key to sketching rational functions. That's right; we're drawing them, and it isn't going to be a total pain in the butt. Believe it.
We've heard it all before. "We can't divide by zero." "We can't have a denominator equal to zero." "A rational function is undefined if the denominator is zero." "If you keep making faces like that, it'll stick that way."
What actually happens if we go ahead and try to do it anyway? For one, we'll have a face that is great for Halloween but bad for any future job interviews. For the whole "division by zero" thing, we get a vertical asymptote.
A vertical asymptote is a line that a function approaches but never quite touches; think of the function like Charlie Brown, and the vertical asymptote is a football. Division by zero is the Lucy of this analogy, pulling the asymptote away from the function at the last second.
The line x = -2 on this graph is a kind of electric fence perpendicular to the x-axis. It causes a bend in the graph of the rational function. It gets very, very, very close to x = -2 but never touches it. It's not actually part of the graph, it just tells us where our function doesn't go.
This graph is for the function . Check out that denominator: it equals zero when x = -2, just like we were saying. Division by zero causes vertical asymptotes. Sometimes we have to do some hunting for them, though.
Find any vertical asymptotes of .
It's just like Mama always said. Life is like a polynomial; factoring will help us find the zeros.
We've got a twofer here, at x = -1 and 3. The function approaches both of these lines, but doesn't cross either one. We wouldn't want to cross an asymptote either; we've heard some scary rumors.
"Division by zero causes vertical asymptotes." We just said that a minute ago. It isn't quite always true, though. Sorry, but there is one big exception to this rule. It's what happens when division by zero cancels out of a rational expression.
This is the graph of . Nothing appears wrong with this, until we realize that x = -1 and 1 both cause the denominator to equal zero. We only have one vertical asymptote, though. Is everything we thought a lie?
Not quite; if we look really close at x = -1, we see a hole in the graph, called a point of discontinuity. The line just skips over -1, so the line isn't continuous at that point. It's not as dramatic a discontinuity as a vertical asymptote, though.
In general, we find holes by falling into them. We don't pay a lot of attention to where we're walking. In rational expressions, we find holes by simplifying the expression.
We cancel out the (x + 1), causing the division by zero to "disappear." However, and this is important, it is still actually there. That's what makes the hole appear. That, or playing t-ball in the house.
The expressions and are graphed the same, except that our original expression has that extra hole in the graph at x = -1. If we forget to include the hole in our graph, then we'll fall into it, and we still have the cast on from the last hole we fell into.
Vertical asymptotes are fairly easy to find. Horizontal and slant asymptotes are a bit more complicated, though. Not actually complicated, but they require a little more work. Just warning you ahead of time.
The horizontal asymptote is the value that the rational function approaches as it wings off into the far reaches of the x-axis. It's all about the graph's end behavior as x grows huge either in the positive or the negative direction. The equation of a horizontal asymptote will be "y = some constant number." Just like the equation of any horizontal line. Hint, hint.
We've already seen a few horizontal asymptotes in the previous section, like in our buddy up there.
Horizontal asymptotes aren't nearly as strict as vertical ones. It doesn't matter to them how many times they are crossed over, as long as the graph approaches them in the end.
Now, here's how we find horizontal asymptotes. It's all a matter of degrees. The degree of the polynomial, that is:
• If the degree of the numerator (up top) is smaller than the degree of the denominator (down below), then the horizontal asymptote is the x-axis itself (y = 0).
• If the degree of the numerator (top dog) is equal to the degree of the denominator (down low Joe), then we look at the leading coefficient of each polynomial. Divide them out, and the horizontal asymptote is y = that number.
Let's take one final look at . Each polynomial has degree 1. So, at that point we ignore every term except the leading coefficients. We have , and as expected, our graph has a horizontal asymptote at y = 1.
Here's another one, .
The numerator has degree 1, while the denominator has degree 2. As x gets larger and larger—let's say 1000—check out the first term of each polynomial. We have . We have to squint to see a number that tiny, and it's going to get even smaller as x gets larger. That's why having a bottom-heavy rational expression leads to a horizontal asymptote at y = 0.
What will happen when a rational function is top-heavy by 1 in its degree? Will it tip and tumble downhill? No, even worse: it forces us to use long division.
That's because a rational function with the degree of the numerator exactly 1 larger than the degree of the denominator has a slant asymptote. It's an asymptote that follows some line, y = f(x). We find the line by going through the long division of our polynomial fraction.
Find the equation of any asymptotes of .
We've got to go through long division here.
Our division has a remainder, but we don't care about that. Throw it out like week-old food. The equation of our slant asymptote is y = x + 4.
Don't go celebrating just yet, though. We also need to look for vertical asymptotes and holes. Yes, we can have multiple types of asymptotes, plus points of discontinuity. Don't freak out over it; just remember to check for all of them. Factoring the numerator, we get:
Nothing cancels out, so we just have a vertical asymptote at x = 1.
See how the line approaches but never touches the line y = x + 4? Slant asymptotes are just as touchy as vertical ones. Poke at your own peril.
For any rational function, we need to check for horizontal and slant asymptotes by scoping out the degree of each polynomial. Don't be a creeper, though; just ask to take a look. Unlike vertical asymptotes and holes, we will only find one horizontal or one slant asymptote.
Okay. Time to bust down some rational functions and graph them.
For rational functions, the graph can be divided into "sections". Each section of the graph is usually defined by their asymptotes. Once we know how they behave there, getting the rest of the graph together is as simple as plotting a few extra points. That makes a lot of sense. It should, given how rational these functions are.
Sketch a rough graph of .
Our first rational function to graph. Let's do it right, by starting with the asymptotes. The degrees of the two polynomials are equal, so we have a horizontal asymptote at y = 1. Let's factor to find any vertical ones, or holes.
A hole at x = 1? Hah, missed us. Not that it was trying that hard to get us. We can also confirm that there's a vertical asymptote at x = 4. Maybe it was out to get us? Or maybe we're just being paranoid.
That's all of the asymptote stuff we can find. Next let's find the x- and y-intercepts.
To find the x-intercept, we set the whole equation equal to 0 and solve for x. That sounds pretty awful for us, because we have x in both our numerator and denominator. But salvation is at hand, or at least a trick for making this easier.
y can only equal zero when the numerator is zero. That means we can ignore the denominator when finding the x-intercept.
0 = x2 – 1
(x + 1)(x – 1) = 0
x = -1 and 1
So, we have y = 0 at the points x = -1 and x = 1. We're almost ready to sketch this out. We have plenty of points to the left of our vertical intercept at x = 4, so we want a few to the right. Then we can…wait a second.
x = 1! We have a hole at x = 1. y can't equal zero there, because the function is undefined there; that's what a discontinuity is. We almost fell for you, x = 1, almost.
We do want those points we were talking about, though. Let's go for x = 5 and 6:
Now let's take all that we've learned about this graph and put it on paper. Or your screen, in this case.
The horizontal asymptotes tell us how the graph behaves at very small or very large numbers. In our case, they approach y = 1. The x- and y-intercepts showed us that the graph approached it from below the line; it would dive sharply downwards at the vertical asymptote. Our other two points gave us the shape of the graph on the other side.
And, of course, we can't forget the hole at x = 1. No, we won't forget you, you sneak.
Our general plan of attack when graphing is:
• Find all of the horizontal, vertical, and slant asymptotes.
• Check for holes, and remembering to graph them. Failure will result in falling in a hole.
• Find the x- and y-intercepts. Just use the numerator when working on the x-intercepts.
• Decide if you need more points to see how the graph is behaving in spots. When in doubt, grab more points.
• Sketch the graph out. We don't need to be exact, just get the general movement correct.
And with that, we are done. Smell ya later, Shmoopers.
Oh yeah, the cookie recipe. Here you go. Now we'll smell ya later. Especially if you've been baking cookies. Don't judge us.