Question

what are (x-4)^2/(36)-(y-2)^2/(9)=1 asymptotes?

Answer 1

hyperbola?

Answer 2

yep

Answer 3

\[\huge \frac{(x-4)^2}{36}-\frac{(y-2)^2}{9}=1 \]

Answer 4

according to your equation above, what is the center here?

Answer 5

(4,2)

Answer 6

ok.. so we KNOW that both asymptotes go through this center... all we need are the slopes...
read here, page 3, about the slopes of hyperbolas:
http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet.pdf

Answer 7

then see if you can come up with the equation of the asymptotes of your hyperbola....

Answer 8

since this hyperbola open left and right so equation of asymptotic is
\[\Large y=\pm(\frac{b}{a}) x\]
where b=3
a=6

so
\[\Large y=\pm(\frac{3}{6}) x=\pm 0.5x\]

Answer 9

@SantanaG did you get it ?

Answer 10

yeah thanks @sami-21

Answer 11

Here are the equations of the asymptotes
\[
y-2 =\pm \frac 1 2 (x-4)
\]