Whooo! Mr. Tarrou I got one last video today. Actually, three videos and this being the
third, I still have not completely covered my section of graphing rational functions.
The next video that I will do in a couple of days will tie all of this together and
we will actually make a complete graph. We are going to find slant asymptotes. Ok, we
just got done talking about vertical asymptotes. That is where the denominator equals zero
and it does not factor and cancel out. I just covered a lecture on finding horizontal asymptotes.
If the top degree is larger than the denominator's degree, then there is no horizontal asymptote.
If the degree of the numerator is the same as the denominator's degree, then the horizontal
asymptote will be y equals whatever the leading coefficients are. If the denominator's degree
is larger then the horizontal asymptote is always y equals zero. So there are three basic
rules for finding horizontal asymptotes of polynomials. In my video I did prove or show
how that actually works or is true with the idea of limits as x approaches infinity. However,
slant asymptotes when the degree of the numerator is one greater than the degree of the denominator.
In this case we have a rational equation where the degree on top is two and the degree on
the bottom is just one. So, this graph is going to have two asymptotes, a vertical asymptote
at x equals three because you cannot divide by zero so there is a vertical asymptote at x equals three. But because the top degree
is larger than the bottom degree by one, there is also a slant asymptote. So let's try and
figure that out. If you have watched a video lately that is called "Long Division of Polynomials",
well if have not we are going to do it again! So, that is how you find slant asymptotes.
You find slant asymptotes through division. Now because we are dividing by a binomial
with a degree of one, you can also short cut this one by doing synthetic division. But,
very often with rational functions you will have something a little bit more complicated
on the bottom so there is a fairly good chance you might be doing long division instead of
synthetic division. So, I am going to show long division instead. Plus students are more
comfortable generally with synthetic division, so why not review something that you might
struggle a little bit more with. So, we have got x squared plus x minus six divided by
x minus three. When we are done with this by the way, the slant asymptote is going to
be whatever is up here. This expression is going to the the slant asymptote when set
equal to y. That is why we are doing the long division. So, how do you get x, this first
term.... I am running out of good colors here that are bright, so I will just do this all
in white. How do I get this first term to equal this first term? I multiply it by x.
I am going to place an x up here. I like to stack mine a little bit differently than textbook
do. I hope that is ok with you. X times x is x squared. That looks like division. Now,
x times negative three is negative three x. Then we are going to subtract down as we do
this long division. X minus x is zero, and x minus negative three x is plus, so one plus
three is four x. Ok, how do I get this leading term to equal this leading term? I am going
to multiply by four. Four times x is four x. Four times negative three is negative twelve.
I am going to go ahead and subtract that. We get zero. Negative six minus negative twelve
is six. And ok, there is a remainder there, but this is really what I am concerned with.
We are going to have a slant asymptote of y equals x plus four. Pretty cool uh:) So,
if I want to graph this function and we are not going to do that completely today because
I am running out of time. I have to get going, even though this video now is relatively short.
But, I will give you a setup, a guide of how you would start this. How much time do I have?
We have a lot of time left. Let's see how much we can do. Ok, so... I just erased my
slant asymptote! Sorry about that. We have got a rational function that is undefined
at three, and therefore we have a vertical asymptote at three. We have a rational function
which has a slant asymptote at y equals x plus four. This has got a slope of one and
a y intercept of four. We are crossing, or we have a slant asymptote that crosses the
y axis at four and from there you are going to up and over one. Here we have a slant asymptote.
Ok, so we are only going to do a quick sketch because finding an x intercept on this is
going to be, uh, new material.... I will just go until the bell rings, or my iPod alarm.
Ok, what have we got here? Let's find the y intercept. That is a good point to find
when you are graphing. So, the intercept is found when x equals zero. If I take this function
and I plug in zero, we are going to have zero squared plus zero minus six all over zero
minus three. Negative six divided by negative three is two. This graph is going to cross
the y axis at two. Let's find the x intercept. The x intercept is when y equals zero. When
is this function going to equal zero? When is y going to be zero? What does x have to be to make this
entire function equal zero? Let's talk about fraction for a second. I do not want the denominator
to equal zero. If the numerator is equal to zero, zero divided ten is zero, zero divided
by 5 is zero, zero divided by a hundred is zero, so with rational function when I am
looking for x intercept, when I am trying to get y to equal zero all I really care about
is the numerator. So, I am going to take the top of this fraction and pull it out. What
is going to make x squared plus x minus six equal to zero? That is going to be x plus
three times x minus two. So, my x intercepts are negative three and two. Negative three
and two. Even though I was a little reluctant to give you a decent sketch of this graph,
it is not really taking that much time. This graph is going to come up from the bottom
of this asymptote. I know it on the bottom because these points are on the bottom. It
going to come up through here, pass through this y intercept, and then come down against
this vertical asymptote that we got from the denominator which is x equals three. Now,
what happens on the other side of the vertical asymptote? The fact that there is no other
x intercepts gives me the idea that it is probably up here. But, I am not going to know
that unless I take the time to do a t-table and try to fill in the part of the graph that
we don't yet understand. Now remember, I just did the y intercept and the x intercept. So,
what is happening to the right of the vertical asymptote of x equals three? Well let's plug
in x equals four and see what we get. I am just doing like a t-table. So, f of four is
equal to four squared plus four minus six all over four minus three. I should be using
parenthesis. That is going to be sixteen plus four minus six over four minus three which
is one. That is twenty and twenty minus six is fourteen. At x equals four we are all the
way up to fourteen. Now, you really should have more than one point to make a graph,
so let's try five and hope the numbers are not too ridiculously big. That should be enough
to let us see what is going to happen. At x equals five we get five squared plus five
minus six all over five minus three. That is twenty-five plus five minus six all over
two. This is thirty, now it is twenty-four, and twenty-four divided by two is twelve.
So there is a little bit of a drop. But I will tell you from experience, this is going
to come up and then approach and slide up the top of that asymptote. Over here the graph
stays at the bottom of the slant asymptote. Really, there needs to be a little more information
to make a good graph. Is it symmetric to the y axis which means that it an even function?
Does it have rotational symmetry to the origin because it is odd? But uh, that is a little
bit more than I planned on doing in this video...but there you go. If your top degree is larger
by one, then you have a slant asymptote. You find it with long division. Vertical asymptotes
come usually when from when the denominator is equal to zero. If the denominator is equal
to zero, but sometimes that can give you a hole in the graph. The top degree is larger
so there no horizontal asymptote. And I found the x intercepts, the y intercepts, I found
a couple of other points. This was just supposed to be for slant asymptotes. I hope I helped
you out some. There will be a video where we do a bunch of these from start to finish,
right from scratch, getting all the important points, talk about all the asymptotes again.
Alright! I am out of here. Go Do Your Homework.