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Graphs with Holes - Problem 3
Let's graph another function with a hole. Here we have a really complicated looking rational function, a cubical over a cubic. The first thing I want to do is factor this. So this whole thing will equal, let's get the numerator there's a common factor of x and that's going to give me -x², actually let me plot a -x. It will give me a positive x² minus 1. And in the denominator I want to factor this as well. Now this requires a trick, it's called factoring by grouping and you're going to factor each pair of terms separately.
I'm noticing an x plus 2 factor here, so let me factor in x² out of the first two. I get x² times x plus 2, and then I'll factor a -1 out of this. -1 leaves an x and a plus 2, -x minus 2. And then you'll notice that we've got x² in the numerator and we've got a factor of x² minus 1 in the denominator. This means I'm going to have not one, but two holes, because this represents 2 linear factors that are common to the numerator and denominator. We've got get x minus 1 and x plus 1 over x minus 1, x plus 1, x plus 2. So these factors will cancel and we'll be left with -x over x plus 2, x not equal to plus or minus 1.
Now this graph is going to be a transformation of 1 over x. It's just a hyperbola and it's going to have two holes in it. When we're graphing functions like these, the first thing we want to do is identify the asymptotes. Now there's going to be a vertical asymptote at x equals -2 and there will also be a horizontal asymptote right? The degree of the numerator is the same as the degree of the denominator, so we find the horizontal asymptote by looking at the leading coefficients. Think of this as -1 times x and 1 times x down here. We have -1 over 1, y equals -1. So those are our two asymptotes; y equals -1 and x equals -2. Let's graph those.
Y equals -1 goes right here, x equals -2 goes right here and we should also find the location of the holes, and that's easy enough to do. We know that the x coordinates plus or minus 1. So the holes, when x is +1 we can just plug in here and get -1 times 1, -1, 1 plus 2, 3, -1/3. And when x is -1, we get -1 times -1, +1, and then -1 plus 2, 1. 1 over 1 is 1, so these are our two holes; 1,-1/3, -1,1. So -1,1 that's a hole and 1,-1/3 right about there.
Now what we need to do is plot a few more points and we'll be able to flush out the rest of our graph. Let's plot a point, since we've got these holes to work with, let's plot a point for x equals 0. Now it's much easier to plug into this equation. If I plug in I get 0 over 0 plus 2, 0. So the graph passes through the origin and I think I'm going to get something like this
Remember it's a transformation of y equals 1 over x and you know what kind of shape to expect. It's kind of hyperbola shape. And then for the other side, I would plot maybe -3 and -4 let's try those. For -3 I get -1 times -3, or 3, and then -3 plus 2, -1. 3 over -1, -3, so (-3,-3). And then for -4, I get -1 times -4 or positive 4, -4 plus 2, -2, 4 over -2, -2 so I get (-4,-2) and you get a graph that looks something like this, that's it.
Look how complicated it looked at the beginning, but when the factors cancel, these holes actually make the graph much easier to make. So remember your holes. Whenever you get a common factor in the numerator and denominator that cancels, it gives you a hole in the graph. And whenever you get a factor in the denominator that gives you a 0, that isn't cancelled by the numerator, you get that vertical asymptote. So that's our final graph.