SOLUTION: I do not understand hyperbolas, so if someone would be able to help me answer these questions and explain how to do them it would be very much appreciated. A hyperbola that passe

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Question 642832: I do not understand hyperbolas, so if someone would be able to help me answer these questions and explain how to do them it would be very much appreciated.
A hyperbola that passes through the point (1, 3) has horizontal asymptote y = -1 and vertical asymptote x = 2.
(a) Find its equation.
Hint: The equation of a hyperbola can be written in the form y = (c/ x - a)+ b

(b) Use your equation to find the x- and y-intercepts.

(c) Draw the graph of your hyperbola clearly labelling its x- and y-intercepts, and asymptotes. You may draw your graph by hand or with computer software.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Some hyperbolas have a horizontal major axis. Some have a vertical major axis. Some have a major axis with a slope of 1 or -1. And then there are those with major axes having other slopes. Only the ones with slopes of 1 or -1 can be written in the form: y = c/(x-a)+b. This is so because this is the equation of a function (only one y value for each x value). And only hyperbolas with a major axis with a slope of 1 or -1 will pass the vertical line test for a function. (If this is not clear to you, try drawing some hyperbolas and try the vertical line test on them.).

So our hyperbola has a major axis with a slope of 1 or -1. The equation for that kind of hyperbola is:
%28x-h%29%28y-k%29+=+c
where (h, k) is the center of the hyperbola and "c" is just some non-zero constant. If c > 0 then the slope of the major axis is 1 and if c < 0 then the slope of the major axis is -1.

The center of a hyperbola is where the two asymptotes intersect. With asymptotes of y = -1 and x = 2, the intersection, and therefore our center, is (2, -1). So now our equation is:
%28x-%282%29%29%28y-%28-1%29%29+=+c
which simplifies to:
%28x-2%29%28y%2B1%29+=+c

To find "c", we just use the point we have been told is on the hyperbola, (1, 3). If it is on the hyperbola, then its coordinates must fit the equation:
%28%281%29-2%29%28%283%29%2B1%29+=+c
which simplifies as follows:
%28-1%29%284%29+=+c
-4+=+c
This makes our equation:
%28x-2%29%28y%2B1%29+=+-4
(Since this form is easier to work with than the y+=+c%2F%28x-a%29%2Bb form, we will postpone converting it until later.)

For the x-intercept we use y = 0:
%28x-2%29%28%280%29%2B1%29+=+-4
%28x-2%29%281%29+=+-4
x-2+=+-4
x+=+-2
So the x-intercept is (-2, 0)
For the y-intercept we use x = 0:
%28%280%29-2%29%28y%2B1%29+=+-4
%28-2%29%28y%2B1%29+=+-4
-2y-2+=+-4
-2y+=+-2
y+=+1
So the y-intercept is (0, 1)

Since our "c" is negative, this hyperbola has a major axis with a slope of -1. This, plus asymptotes and the three points we have (the two intercepts and the given point of (1, 3)), may be enough for you to graph the equation. If not, then pick some other x's (or y's) and use the equation to find the corresponding y's (or x's). This will give you additional points to plot. Repeat finding and plotting points until you can see the path the graph follows.

Finally, we will transform
%28x-2%29%28y%2B1%29+=+-4
into y+=+c%2F%28x-a%29%2Bb form. Dividing both sides by x-2:
y%2B1+=+%28-4%29%2F%28x-2%29
Adding -1 to each side:
y+=+%28-4%29%2F%28x-2%29+%2B+%28-1%29