Instead of doing yours for you, I'll do one
exactly like it, which you can use as a model:




 
, , 
 
, so 
 
, so 
 
The center (h,k) = (-2,7)
 
We start out plotting the center C(h,k) = C(-2,7)
 

 
Next we draw the left semi-transverse axis,
which is a segment a=5 units long horizontally 
left from the center.  This semi-transverse
axis ends up at one of the two vertices (-7,7).
We'll call it V1(-7,7):
 

 
Next we draw the right semi-transverse axis,
which is a segment a=5 units long horizontally 
right from the center. This other semi-transverse
axis ends up at the other vertex (3,7).
We'll call it V2(3,7):
 

 
That's the whole transverse ("trans"="across",
"verse"="vertices", the line going across from
one vertex to the other. It is 2a in length,
so the length of the transverse axis is 2a=2(5)=10
 
Next we draw the upper semi-conjugate axis,
which is a segment b=8 units long verically 
upward from the center.  This semi-conjugate
axis ends up at (-2,15).
 

 
Next we draw the lower semi-conjugate axis,
which is a segment b=8 units long verically 
downward from the center.  This semi-conjugate
axis ends up at (-2,-1). 
 

 
That's the complete conjugate axis. It is 2b in length,
so the length of the transverse axis is 2b=2(8)=16
 
Next we draw the defining rectangle which has the
ends of the transverse and conjugate axes as midpoints
of its sides:
 

 
Next we draw and extend the two diagonals of this defining
rectangle:
 

 
Now we can sketch in the hyperbola:
 


The foci are points inside the hyperbola, which are the distance c
from the center, where c is calculated by

  (just like the Pythagorean theorem, from whence
it comes):







So the two foci are  units 
right and left of the center, which is
(h,k) = (-2,7)

Therefore the foci are:

(,7) and (,7) 

They are approximately the points:

(-11.4,7) and (7.4,7).  I won't bother plotting
them.

All that's left to do is find the equations of the two asymptotes.
Their slopes are ± or ±
 
The asymptote that has slope  goes through the center
C(-2,7), so its equation is found using the point-slope
formula:
 



Multiply through by 5



 
The asymptote that has slope  goes through the center
C(-2,7), so its equation is found using the point-slope
formula:
 



Multiply through by 5



 -----------------------------------------------
Edwin