HELLO, Mr. Tarrou
I have done a lot of videos both vertical asymptotes, horizontal asymptotes, slant asymptotes.
And on that slant asymptotes video I ended making a full graph from start to finish.
I want to some more of that today with rational functions. (polynomial divided by polynomial)
For our first example we are going to take a look at y equals x minus 2 over x minus 3.
Ok ...so...
Lets look for vertical asymptotes.
The first thing you want to do for vertical asymptotes...
That is like questions we looked at in chapter one of our PreCalculus book where you were being asked about the domain of the function.
You cannot divide by zero.
So vertical asymptotes will occur most of the time where the denominator equals zero.
In this case we have x minus 3 in the denominator that cannot equal 0.
This is also going to answer "What is the domain of this graph?"
That means that x cannot equal 3.
So, we are going to have a vertical asymptote for this rational function at x equals 3.
Now in the next example you are going to see, if you have a zero in your denominator and it cancels out, that is going to make a hole in the graph.
Otherwise whenever that denominator can equal zero, and it cannot be cancelled out, that is going to yield a vertical asymptote.
Ok, horizontal asymptotes.
In my video about horizontal asymptotes I showed you how to truly find them using limits.
But then we looked at three basic rules.
If the degree of the numerator is larger, then there is no horizontal asymptote.
If the degree in the numerator and denominator are equal, the horizontal asymptote is y equals the numerator's degree divided by the denominator's degree.
If the denominators degree is larger then your horizontal asymptote is y = 0.
Well, our numerator has a degree of one and our denominator has a degree of one, and since these degrees match our horizontal asymptote is
is y equals the leading coefficient of the top divided by the leading coefficient of the denominator.
That is going to be y equals one over one.
or one. So our horizontal asymptote, because the degree of the numerator equals the degree of the denominator...they are both one, is going to be y equals one.
Ok, so we have our vertical and horizontal asymptotes.
There is not a slant asymptote because we have already determined that there is a horizontal asymptote.
Slant asymptotes occur when the degree of the numerator is one larger than the degree of the denominator.
You find the equation of the slant asymptote using division. You can check out my other video lesson:)
What else does this graph have going on? What is the x intercept? You find the x intercept by letting y equal zero.
That is going to give us zero equals x minus two over x minus three.
Not forget we are dealing with a fraction. How do you get a fraction to equal zero?
If the numerator equals zero then the entire fraction equals zero.
If the denominator equals zero, then the fraction is undefined. That is the question we asked when looking for vertical asymptotes or domain.
So when you are dealing with rational functions and you want to find where it crosses the x axis, crossing the x axis means that y must be zero, then all you need to worry about is the numerator.
zero equals x minus two.
You would get the same equation if you multiplied both sides by x minus three, it would disappear anyway.
Add two to both sides and my x intercept is x equals two.
What else can we figure out about this rational function? Lets see if we can figure out what the y intercept is.
You find the y intercept by letting x equal zero. This is usually a relatively easy number to find,
because of how easily the algebra works out.
Our y intercept when we let x equal zero is going to be zero minus two over zero minus three. I have just substituted this zero in for x.
Negative two divided by negative 3 reduces to be positive two thirds.
So my y intercept is two thirds.
Another thing to check which would be nice for graphing is to test whether this function is even or odd.
Even or Odd. You do that by plugging in negative x and seeing if all the signs are the same or if you get an opposite answer(for polynomials all the signs will change).
We are going to write negative x minus two over negative x minus three.
This gives us negative x minus two over negative x minus three.
First of all, is it even. If I plug in negative x and I get right back to the equation that I started with then the function is even.
I don't have exactly the same function. Some of the signs changed and some of them did not.
Why would I care if the function is even for sketching?
If a function is even then it is symmetric to the y axis. So if I am sketching I can make sure the graph is correct on the left side of the y axis...
...did I just say it was symmetric to the x axis, I hope not...
Even functions are symmetric to the y axis so the graph to the left of the y axis and to the right of the y axis will look the same.
My function is not the same as what I started with so this is not even.
Why would I care if the function is odd. If the graph has 180 degree rotational symmetry about the origin, what we calls symmetric to the origin, it is odd.
Knowing this would allow us to make a better sketcher...quicker. This gives us a better idea of what it looks like.
In a polynomial function that is odd, all of the signs change. Two of my signs changed, the ones in front of the x but my constant signs changed the same.
This function is neither even nor odd.
So, I thing we are ready to draw a picture.
We have a graph with a vertical asymptote at x equals three.
Here is our vertical asymptote at x equals three.
We have a horizontal asymptote at y equals one.
The blue and yellow lines again are the asymptotes.
Our x intercept is two.
We have our y intercept is two thirds.
And that is about all we know about this graph so far.
I would like to find a few more points.
A graph may cross a horizontal asymptote occasionally but it will never cross a vertical asymptote.
So this graph is coming through this point and it will curve down and go towards negative infinity as we approach three from the left.
I can check this by doing, lets say, x...y...use and x value of point five and plugging it in here.
But I know this is going to curve down, so lets use negative one.
Lets find out what happens to the right of the vertical asymptotes.
We are going to use the x values of 4,5, and 6.
Negative one minus two is negative 3 and negative one minus three is negative four.
This equals three fourths.
At negative one we are at three fourths.
So we are a little bit closer to one and this graph is going to take off to the left getting closer to the horizontal asymptote y equals one.
It is just going keep coming closer to y equals one from the bottom.
At four, we have four minus two equals two and four minus three equals one, so we have two.
So at four we have a y value of two.
At five, and I think that is all we will need, at five we have five minus two over five minus three which is three over two.
This is 1.5 so at 5 we are at 1.5
This graph is going through these points and will continue to get closer and closer to this horizontal asymptote of y equals one as x approaches infinity.
As I get closer to the vertical asymptote x equals 3 from the right, it is going to continue to go up to positive infinity.
So there you go. The white part of this drawing is the graph of y equals x minus 2 over x minus three.
So, graphing rational functions. Lets try to do one more.
Ok, I am not going to have time for another example so let me go ahead and start another video.
Go to the next part and we are going to graph a rational function that has both a hole in it and a vertical asymptote.
Don't forget to look up slant asymptotes. There is a lesson with a complete graph/example where I show you how to find that slant asymptote, find the x and y intercepts, and some additional points in the graph.
So you will have three complete examples of graphing rational functions
ALRIGHT, be back in a second.