University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Graphing a transformed hyperbola combines the skills of graphing hyperbolas and graphing transformations. With hyperbola graphs, we use the formula a^2 + b^2 = c^2 to determine the foci and y= + or - (a/b)x to determine the asymptotes. When transforming hyperbola graphs, we find the center of the graph and then graph accordingly.
Graphing a transformed hyperbola basically puts together everything that we know about transformations and hyperbolas okay? So the first thing we want to look at is how these minuses affect the curve and just like with a circle or just like with an ellipse what we're looking for is it's always x minus the x coordinate of the center y minus the y coordinate of the center either way you can think about it is what value can we put in that will give us 0 and that will be the coordinate for the center, so looking here x-3 tells us that 3 is our x coordinate for our center y+1 y minus -1 that tells us -1 is our y coordinate so we end up with 3, -1 as our center okay?
From there, all we have to do is view our center as the origin okay so just before we did everything off of the origin over made our silly box all that stuff we're doing the same exact thing now but instead of going off of the origin we're just going to go off of this point okay? So what this tells me is the x is first which tells us it's a horizontal hyperbola and the x the term underneath the x is 4 which tells us that this is 2 squared so our sort of x radius our x dimension is going to be 2 in each direction so all I need to do is draw out 2 in either direction and that marks my "x radius" in this case it's going to be the transverse axis because this is the direction of the hyperbola okay? Doing the same I think for not for the y term 9, 9 is 3 squared so this tells us we're going to go up 3 and down 3 from the center so from the center up 3, from the center down 3 we were at 1 so that takes us to -4 and from here we are able to draw our box okay?
So we basically oh! The first we want to do is connect where these points would actually meet up so we take these two points and take this one over take this one straight up that gives us a point another point third point and a fourth point so that makes up our box if we were to draw. You can either connect the dots if you want to or the main reason we do all this is to find our asymptotes okay? so basically all we want to do is connect the opposite corners of our rectangle, and that line is horribly off I apologize, hopefully you at least get the idea of what's going on.
And now we just want to figure out where the graph goes okay? So the endpoints of our transverse axis are actually the turning points for a graph where that graph turns from coming one way to go in the other so I know that the x is first so this is why I want my vertices, this point is on the graph as is this point over here and then just filling in the blanks we know that the graph has to get close to these asymptotes and go through this point, so we have that one half over here and the other half ends up over here.
It's not perfection obviously but I hopefully you got a good idea of how to graph a hyperbola that has transformed. Basically find the center and then do everything you did when it was centered around the origin but just off of that new center point.
Unit
Conic Sections