Graphing Rational Functions, n>m - Problem 2 1,970 views

I want to graph another rational function where the degree of the numerator is bigger than the degree of the denominator. Now here, usually the first thing I like to do, is find the x intercepts and the asymptotes. But I’m going to need the numerator to be factored in order to that, so let’s factor the numerator.

We’ve got an x² here, so the factors are going to start with and x and we have a 3 here. So I’m hoping that a 1 and a 3 will work. It looks like it will because 3 minus 1 is 2, so let’s try -3 and plus 1. That will definitely give me my -3 and it will give me -3x plus x which is -2, so it does work. And so that’s my factored form. And remember that x intercepts come for the zeros of the numerator, so I have an intercept at -1,0 and at 3,0 and lets plot those right away. -1,0 is right here and 3,0 is over here.

Now let’s find the asymptotes. Where you get a vertical asymptote when x equals 0 because you got the rational by 0, and the non-vertical asymptote, the oblique asymptote if it exists will find it by polynomial division. Remember that a rational function does not have a horizontal asymptote if the numerator has a higher degree than the denominator. So we have to look for an oblique asymptote instead.

So we get x² minus 2x minus 3 divided by 2x. So what do we need to multiply by 2x to get x²? 1/2x. I multiply through and I get x² and then I subtract, -2x. What do I multiply by 2x to get -2x? -1. Multiply 2 to get -2x change signs and add and my remainder is going to be -3.

So what this means is, that my function can actually be written y equals this thing, the quotient, 1/2x minus 1 plus the remainder of -3 over the divisor 2x.

Now if you look at this part of the function, you can see that as x gets very large, this thing is going to diminish away to 0 and you’ll be left with this line. That means that the graph is going to approach this line of slope 1/2 and y intercept -1. So that’s going to be our oblique asymptote. And we’ll plot that; y equals 1/2 x minus 1. Well minus 1 is here. A slope of 1/2 but it's going to go up 1 over 2, so it will pass through here. So let me draw that line. It looks something like that.

That’s y equals 1/2 x minus 1. And don’t forget we also have our vertical asymptote x equals 0 and that’s just the y axis. So the vertical asymptote in the intercepts divide the plane up into regions, let’s plot some points. How about -2?

So what happens when x is -2? If I look up here, I get -2 plus 1, -1, -2 minus 3 -5 so -1 times -5 over 2 times -2, -4. So it's 5 over -4, -5/4. (-2 -5/4) that’s just a quarter less than -1, so it will be down here. That’s actually enough that we can probably graph the left half. Remember that we have a vertical asymptote here, so it's going to have to go through and up approach the y axis asymptotically, and then down and approach the oblique asymptote.

What happens over here? Let’s plot a point at x equals 1, see what happens. X equals 1 we get 1 plus 1, 2 and 1 minus 3, -2. And the denominator we get 2 times 1, 2 the 2’s cancel we get -2 so (1,-2).

And actually put it a little bit lower. That might be enough to finish the graph. It's going to approach this asymptote and it will have to go down and approach the vertical asymptote, so something like this. And that’s all there is to it.

Just find your intercepts and your asymptotes and then plot a few points just to flash out your graph.