Hyperbola: Graph from equation: 1st quadrant: Find the standard equation and draw the graph for

The form of the standard equation is .

First let's group the x and y terms:

.

Next we factor as follows:

Now we complete the squares by adding and subtracting within the parentheses:

.

Next move the subtracted terms outside the parenthese (and remember the factor)

Now move the "outside"constants to the right hand side of the equation and write the parentheses as squares:

Since we need a "1" on the right-hand side in order to compare with the general equation, we now divide by 64 to get:

After canceling, we have

Now we can conclude that h =2, k = 3, a is 4 and b is 2.

The standard graph has its center at (h, k). Here that is (2, 3). The vertices of the hyperbola are "a" units to the right and left of the center, or at x = -2 and x = 6.

The asymptotes go through the center with slopes +b/a and -b/a. In this case the slopes are 2/4 and -2/4.

We can check our results by substituting the x-coordinate of a vertex into the original equation and determining the corresponding y-coordinate. It should agree with the value of k, the y-coordinate of the center and the vertices. Using x = 6, we get

, or

, or

, or

, and the value of y agrees with the y-coordinate of the vertex, k = 3. You can check in a similar way with the x-coordinate of the other vertex: x = -2.

Draw your graph. When complete, check it.