 We categorize conics by their eccentricity. In a parabola, our eccentricity is equal to 1. And what's important to realize here is that this actually means there's only one parabola. All parabolas look identical, and the only difference is how they're oriented. For an ellipse, the eccentricity is between 0 and 1, and this means there are different types of ellipses, but we can graph them by finding the endpoints of the major and minor axes. And in a hyperbola, the eccentricity is greater than 1, and we can graph these by finding the vertex, but what then? The problem is there's an infinite number of hyperbolas with the same vertex, and so we need a way to distinguish between them. To solve this geometric problem, we'll do a little bit of algebra. So let's take one of the equations for a hyperbola, and I'll have this one. And let's solve this for y. Now, remember, one of the things we are interested in is long-term behavior. And if x is very large, then it dominates this radicand, and so we can say that y is approximately the square root of just the first term of the radicand, which reduces to b divided by a times x. And another way of looking at this is that y gets close to the line b divided by ax. In other words, we have a pair of asymptotes. And so this gives us a theorem, but again, don't memorize formulas. Understand concepts. We got the asymptotes by solving for y and considering what happens for x very large. So let's find the asymptotes, then graph this hyperbola. So the first thing to notice is that this is a transformation of y squared divided by 5 squared minus x squared divided by 4 squared equal to 1. So to find the asymptotes, we observe the following if we solve for y. And again, while this is exact for very large values of x, the radicand becomes very similar to the first term here. And so y is approximately plus or minus 5 fourths x. So the hyperbola approaches the asymptotes y equals plus or minus 5 fourths x. So we'll draw those. But since they're not actually part of the graph, we'll use dotted lines. Now we can continue as before. The least x squared or y squared can be is 0. But if y squared is 0, we won't be able to solve the equation. If we let x squared be 0, we find. So we know that 0 5 and 0 negative 5 are on the graph. And now we can sketch the hyperbola through these points with the given asymptotes. Now we have the graph of y squared divided by 5 squared minus x squared divided by 4 squared equal to 1. But that's not quite the graph we want. So we have to apply a little transformation. And so the graph we want will be this graph shifted one unit to the left, then shifted up three units. And this affects the asymptotes in exactly the same way, shifting them one left and up three units. And so the equation for the asymptotes will be.