 Let's move on. We're going to try to graph this thing and we're going to need to find some different things about our function. The first thing you would want to do is find the y-intercept. In other words, let x be zero. And a good way to do that, because all the other things we want to do is factor first. These nice things happen if you factor first. So on the top we would have x times x plus three. And on the bottom we'd have x minus three and x plus one to give us a minus two. So if we let everything be zero in here, if I let this be zero and this be zero plus three and multiply, that's going to give me zero on top. Zero divided by anything is just going to be zero. So we would have, so we're going to have a y-intercept at zero zero. Again, zero divided by anything is zero. So now we want to find the zeros of the numerator. Well that means that x is going to be equal to zero. So there it is. And it also means that x plus three is going to be equal to zero. So x is going to be equal to negative three. So my two zeros are at zero, which we figured from the y-intercept, and negative three. Now we want to find the zeros of the denominator. Well that's going to be x minus three equals zero. And that will give us x equals three. And then we have x plus one equals zero, and that's going to be x equals negative one. Now what we found here was the y-intercept. When we found the zeros of the numerator, we actually found the x-intercepts. And when we found the zeros of the denominator, what we really found are the vertical asymptotes. And we want to determine if it changes side from one asymptote to the other. Well we have to go back and look at the degrees. And they're both odd. So that means that we will change sign at both vertical asymptotes. Compare the leading terms. So this is what we have here. We're trying to find horizontal asymptote. And this time they are equal. x squared is equal to x squared. Actually it's the twos that are equal to each other. That's what we really care about. When the degrees are the same, then it's going to be y is equal to a over b. And in our case it's just going to be y is equal to one over one or just one. So our last asymptote is going to be at one. So let's just go ahead and draw that in here. Put dotted line in there because that's the horizontal asymptote. And let's go ahead and put in what we learned from the last one. We found out that our vertical asymptotes, this was our horizontal asymptote. And our vertical asymptote is at three and negative one. So let's put those in here. And what else did we find out? We found out the y intercept was zero, zero. And that the zeros of our graph are at negative three and zero. So now we have to continue on where we were. Determinate the grass crosses that horizontal asymptote. So we say that x squared plus three x over x squared minus two x minus three is equal to that horizontal asymptote of one. So we multiply by x squared minus two x minus three so that we can clear the fraction. And when we do that, we have x squared plus three x is equal to that. I'll just leave it since it's in black. This is what we have right here. So let's gather everything to one side if we need to because we've got quadratics. But when I do that, if I take this x squared to this side, they're just going to cancel each other out. So now I have negative two x minus three plus three x. So we want to take this two to the other side. And that will give me negative three over here is equal to adding that will give me five x. And then I have x is equal to negative three fifths. In fact, I'm going to put it on this way negative three fifths. So now we need to see that we have a point here and we know it's going to cross here at negative three fifths, which is somewhere before negative one. I have a point here in this region. I have a point in this region, but I might like to have a point over here on this side just to verify the things I think. So I'm going to choose, let's say four. So x is equal to four. So four, four squared plus three times four all over four squared minus two times four minus three. And we get five point six. So if we're at four here, we're at a little bit more than five sets up here. So we said that our graph has to be in this region. It's got to go through this point because it's a next intercept. So I'm going to draw my graph to look something like this. And then I know that it's switch size. That's what I was expecting because it was an odd degree on that vertical asymptote. And I have to come through that point that I found plus I have to find this intercept. But I also know that I got to get over here to this vertical asymptote because I'm going to switch sides over here. So my graph is going to probably look something like this. And then I come back on the other side of my vertical asymptote and back on top. And it doesn't cross the horizontal asymptote because we only found one place that it crossed it. So it's just stay above that horizontal asymptote. Use the asymptote zeros and multiplicities to construct an equation for the graph and be sure to find a value for a. If we think about the equation, we have the zeros of the numerator. And then we've got the zeros of the denominator. And these are x-intercepts and these are vertical asymptotes. And then we can check the degrees. Remember here, if we're checking the degree here, that means it's even will bounce and odd will cross. And if we're looking at vertical asymptotes, even will not change sign and odd does change sign. And then the only thing we have to do is find that value for a. So let's see what we have. We have vertical asymptote at x equal negative 3 and x equal positive 3. Negative 3 is between that negative 2 and negative 4. And between the 2 and 4 is going to be 3 because my scale is 2. And then the x-intercepts, there's only one. It's at 0, 0. So we have x equals 0. We see that this one has to be an odd because it crosses at that point. And let's go back and look at x equal negative 3. On the left-hand side, it's above the x-axis. And on the right-hand side of that asymptote, it's also above the x-axis. So this one here is going to have to be even. And then positive 3, on the left-hand side of that one, it's in the negative. It's below the x-axis and it jumps up on the other side to be above the x-axis. So this one will have to have an odd degree. And the only other thing we have to worry about is A. So now we have to think about the horizontal asymptote. And the horizontal asymptote is y equals 0 because we don't see a dotted line anywhere. So it must be the x and you can see it's tending toward the x-axis. And that means that the degree of the numerator has to be less than the degree of the denominator, which earlier we called m. So let's see what we can do. And then we have our A and x equals 0. The factor would be x. And on the bottom, we have x equal negative 3. So it would be x plus 3. And that would be an even degree. So it would be squared. And then it would be x minus 3 when x is equal to 3. And that's an odd one. So we'll just leave it x. This is a degree 3, which is bigger than degree 1. So we're good there so far. So good. Now we have to use this negative 2, 2 here is going to plug into our equation. So we have 2 is y equal to A times negative 2 over negative 2 plus 3 squared and negative 2 minus 3. So 2 is equal to negative 2A. And on the bottom here, this is going to be 1 squared or 1. And this is negative 5 times negative 5. Clear the fraction. So we want to multiply both sides by negative 5 to clear that fraction. And those cancel. Now we have negative 10 equal to negative 2A. And A is equal to 5. I'm going to get rid of my A and replace it with 5. And this is the function.