Like what you saw?
Create FREE Account and:
Graphing Rational Functions, n>m - Problem 3
Let’s try a harder example. We are graphing rational functions where the degree of the numerators is bigger than the degree of the denominator. So let’s take a crack at this one.
First thing I like to do is find the x intercepts and the asymptotes and I’m going to need this to be factored. Let’s make that easier. So let’s factor the numerator. Notice there’s a common factor of x² so I’ll pull that out. And that leaves a 4 from this term and an x from this term. And the denominator is a quadratic let’s see if that factors. It starts with and x² so let’s try x and x and it ends with a 10. So 2 and 5, that should work because 5 minus 2 is 3, so let’s try -5 and plus 2. That does give me -10 and -5x plus 2x gives me -3x so that works.
X intercepts; these come from the zeros of the numerator, and the zeros of the numerator is 0 and 4. So 0,0 and 4,0 are my x intercepts and I plot those. Next, vertical asymptotes, they come from the zeros of the denominator. So we get x equals 5, x equals -2. So let’s graph those.
That’s 6, that’s 5, so x equals 5 is right here and -2, half of between 0 and -4, so right there. Those are vertical asymptotes. And now recall that when you have a rational function and the degree of the numerator is bigger the degree of the denominator, you are not going to have a horizontal asymptote but you might have an oblique asymptote. So let’s use polynomial division to find out. I’m going to divide the numerator by the denominator. So -x³ plus 4x² divided by x² minus 3x minus 10.
So what do we multiply by x² to get -x³? -x. Multiply through -x³, 3x², 10x, change signs and add. Get 0, x² minus 10x. What do I have to multiply by x² to get x²? 1. Plus 1, times I get x² minus 3x minus 10, I change signs and add. I get 0 here -7 x and plus 10.
And I know I’m done because this has a degree less than the divisor. So this is going to be my remainder. And the means that I can write my function as y equals, the quotient -x plus 1, plus the remainder -7x plus 10 over the divisor. x² minus 3x minus 10.
Now we know that as x goes to infinity, this thing is going to go to 0, because the degree of the denominator is smaller than the degree of the numerator. This will go to 0. And so the curve will behave more and more like this line the slanted line y equals -x plus 1, that is our oblique asymptote; Y equals -x plus 1. Let’s plot that right now.
Plus 1, 1 is the y intercept of the other oblique asymptote and the slope is -1. So I’ll go down 5 and over 5, that will be another point. I like to pick point that are reasonably far apart to graph my lines. So this is what the oblique asymptote looks like.
Now I’m going to need to plot some points to figure out what happens in this left region, in this middle region and in the right region. So let me start with -4. -4² is 16, 4 minus -4, 8, -4 minus 5 -9, -4 plus 2, -2. So the -2 and the 8 cancel leaving a 4. So I get 64 over -1 times -9 is 9. This is approximately 7, 64 over 9 is a little more than 63 over 9 which is exactly 7. So -4 and approximately 7, that’s 6, that’s 7, about here.
Now I probably don’t need to plot any more points, we have an asymptote on the left and an asymptote on the right. I think the graph is going to look something like this and that was pretty easy. So maybe I should try to graph over here next and see what happens, that might be easy as well.
So let’s try 6. We get 36 times 4 minus 6, -2, 6 minus 5, 1, 6 plus 2, 8. So this is going to cancel a little bit. We’ve got we get -72 on top and an 8 that cancels leaving -9. So this is -9, (6, -9) so that’s 6, -8, -10 -9. So we are in this region here. Well we might have a shape like this, let’s plot one more point just to be sure. And just to speed things up I’ve already calculate the value at 8 and it's minus 128 over 15.
Now 128 over 15 is approximately -8 and 1/2, so we have 8, -8 and 1/2. So we are in here. So it looks like we do have something kind of like what we have in our other regions to the left here, very similar. Now what happens in the middle? Let’s go back to this form of the question. Notice the two numerator factors. This one squared.
If you remember when you were graphing polynomials, when you have a square factor, the graph hits the axis and bounces off. The same this is going to happen here. We will either have it bouncing like this or like this.
One of those two. So all we have to do is plot one or maybe two points to find out. Let’s plot x equals 3 and I take the liberty of already calculating that. It's -9/10. So I’ll plot that 2, 3, -2, -1, -9/10 is really close to -1. So about here. It looks like it’s going to bounce off, come down to here, go back up and approach this line asymptotically and then do the opposite down here. That’s a tough one but it’s pretty easy if you just remember asymptotes, intercepts. And if you need to remember that when you have repeated factors in the numerator, that sometimes means you get that sort of bouncing effect that you get like when you are graphing polynomials. The graph hits the axis and bounces off, and that’s our graph.