This is a hyperbola because the  and the  term have opposite
signs.



Get it like this:



Put parentheses around the first two terms:



Factor -1 out of the last two terms on the left



Take half of -8, get -4, square it get +16, add +16 inside the 
first parentheses and add +16 to the right side:



Take half of -2, get -1, square it get +1, add +1 inside the second
parentheses, but because of the - in front of the second parentheses
on the left we add -1 to the right side:



Factor the two parentheses as perfect squares, Combine the terms on the
right.



Get a 1 on the right side by dividing every term through by 2





Compare that to



and since the term in x is the positive one, we know the hyperbola 
opens right and left. (As we know the x-axis goes right and left, a
way to remember it).

Comparing further, center = (h,k) = (4,1)

We plot the center



Comparing further:

 so 

Length of semi-transverse axis = a = 

So we draw the transverse axis  about 1.4 units
to the right and to the left of the center:



The ends of the transverse axis are the vertices. We subtract "a" 
from "h" to get the x-coordinate of the left vertex, so the left
vertex is 

(4-,1).  

We add "a" to "h" to get the x-coordinate of the right vertex, so 
the right vertex is V(4+,1).  They have the same 
y-coordinate as the center. 

 so 

Length of semi-conjugate axis = b = 

So we draw the conjugate axis  about 1.4 units
above and below the center:



Next we draw the defining rectangle around the two axes:


 
We draw the asymptotes by drawing the extended diagonals of the
defining rectangle:



These asymptotes have slopes 

The go through the center (4,1) so their equations are gotten using
the point-slope formula:



Finding the equation of the asymptote with the positive slope





Finding the equation of the asymptote with the begative slope





Finally we sketch in the hyperbola:



Edwin