Current time:0:00Total duration:8:20

# Graphing rational functionsÂ 2

CCSS Math: HSF.IF.C.7, HSF.IF.C.7d

## Video transcript

Let's do a couple more examples graphing rational functions. So let's say I have y is equal
to 2x over x plus 1. So the first thing we might
want to do is identify our horizontal asymptotes,
if there are any. And I said before, all you
have to do is look at the highest degree term in the
numerator and the denominator. The highest degree term here,
there's only one term. It is 2x. And the highest degree
term here is x. They're both first-degree
terms. So you can say that as x
approaches infinity, y is going to be-- as x gets
super-large values, these two terms are going to dominate. This isn't going to
matter so much. So then our expression, then y
is going to be approximately equal to 2x over x, which
is just equal to 2. That actually would
also be true as x approaches negative infinity. So as x gets really large or
super-negative, this is going to approach 2. This term won't matter much. So let's graph that horizontal
asymptote. So it's y is equal to 2. Let's graph it. So this is our horizontal
asymptote right there. y is equal to 2, that right
there-- let me write it down-- horizontal asymptote. That is what our graph
approaches but never quite touches as we get to more and
more positive values of x or more and more negative
values of x. Now, do we have any vertical
asymptotes here? Well, sure. We have when x is equal to
negative 1, this equation or this function is undefined. So we say y undefined when
x is equal to negative 1. That's definitely true because
when x is equal to negative 1, the denominator becomes zero. We don't know what 1/0 is. It's not defined. And this is a vertical asymptote
because the x doesn't cancel out. The x plus 1-- sorry--
doesn't cancel out with something else. Let me give you a quick
example right here. Let's say I have the equation
y is equal to x plus 1 over x plus 1. In this circumstance, you might
say, hey, when x is equal to negative 1, my
graph is undefined. And you would be right because
if you put a negative 1 here, you get a 0 down here. In fact, you'll also
get a 0 on top. You'll get a 0 over a 0. It's undefined. But as you can see, if you
assume that x does not equal negative 1, if you assume that
this term and that term are not equal to zero, you can
divide the numerator and the denominator by x plus 1, or you
could say, well, that over that, if it was anything
else over itself, it would be equal to 1. You would say this would be
equal to 1 when x does not equal negative 1 or when these
terms don't equal zero. It equals 0/0, which we don't
know what that is, when x is equal to negative 1. So in this situation,
you would not have a vertical asymptote. So this graph right here,
no vertical asymptote. And actually, you're probably
curious, what does this graph look like? I'll take a little aside here
to draw it for you. This graph right here, if I had
to graph this right there, what this would be is this would
be y is equal to 1 for all the values except for x
is equal to negative 1. So in this situation the graph,
it would be y is equal to 1 everywhere, except for
y is equal to negative 1. And y is equal to negative
1, it's undefined. So we actually have
a hole there. We actually draw a little circle
around there, a little hollowed-out circle, so that we
don't know what y is when x is equal to negative 1. So this looks like
that right there. It looks like that
horizontal line. No vertical asymptote. And that's because this term and
that term cancel out when they're not equal to zero,
when x is not equal to negative 1. So when your identifying
vertical asymptotes-- let me clear this out a little bit. when you're identifying vertical
asymptotes, you want to be sure that this expression
right here isn't canceling out with something
in the numerator. And in this case, it's not. In this case, it did,
so you don't have a vertical asymptote. In this case, you aren't
canceling out, so this will define a vertical asymptote. x is equal to negative 1 is a
vertical asymptote for this graph right here. So x is equal to negative 1--
let me draw the vertical asymptote-- will
look like that. And then to figure out what the
graph is doing, we could try out a couple of values. So what happens when
x is equal to 0? So when x is equal to 0 we have
2 times 0, which is 0 over 0 plus 1. So it's 0/1, which is 0. So the point 0, 0
is on our curve. What happens when
x is equal to 1? We have 2 times 1, which
is 2 over 1 plus 1. So it's 2/2. So it's 1, 1 is also
on our curve. So that's on our curve
right there. So we could keep plotting
points, but the curve is going to look something like this. It looks like it's going
approach negative infinity as it approaches the vertical
asymptote from the right. So as you go this way it, goes
to negative infinity. And then it'll approach our
horizontal asymptote from the negative direction. So it's going to look
something like that. And then, let's see, what
happens when x is equal to-- let me do this in
a darker color. I'll do it in this red color. What happens when x is
equal to negative 2? We have negative 2 times
2 is negative 4. And then we have negative-- so
it's going to be negative 4 over negative 2 plus 1,
which is negative 1, which is just 4. So it's just equal
to negative 2, 4. So negative 2-- 1, 2, 3, 4. Negative 2, 4 is on our line. And what about-- well, let's
just do one more point. What about negative 3? So the point negative 3-- on the
numerator, we're going to get 2 times negative 3 is
negative 6 over negative 3 plus 1, which is negative 2. Negative 6 over negative
2 is positive 3. So negative 3, 3. 1, 2, 3. 1, 2, 3. So that's also there. So the graph is going to look
something like that. So as we approach negative
infinity, we're going to approach our horizontal
asymptote from above. As we approach negative 1, x is
equal to negative 1, we're going to pop up to positive
infinity. So let's verify that once again
this is indeed the graph of our equation. Let's get our graphing
calculator out. We're going to define y as 2x
divided by x plus 1 is equal to-- delete all of that out--
and then we want to graph it. And there we go. It looks just like
what we drew. And that vertical asymptote, it
connected the dots, but we know that it's not
defined there. It just tried to connect
the super-positive value all the way down. Because it's just trying out--
all the graphing calculator's doing is actually just making
a very detailed table of values and then just connecting
all the dots. So it doesn't know that this
is an asymptote, so it actually tried to connect
the dots. But there should be no
connection right there. Hopefully, you found this
example useful.