How do you find the domain and range of #(y^2)(x-4)=8#?

1 Answer
Jan 23, 2018

Given: #(y^2)(x-4)=8#

#y^2 = 8/(x-4)#

#y = +-(2sqrt2)/sqrt(x-4)#

Separate into two equations:

#y = -(2sqrt2)/sqrt(x-4)# and #y = (2sqrt2)/sqrt(x-4)#

In both cases, the expression under the radical cannot be negative and it must be greater than 0:

#x -4 > 0#

#x > 4 larr# this is the domain for both equations.

Find the range for the first equation:

#lim_(x to 4) -(2sqrt2)/sqrt(x-4) = -oo#

#lim_(x to oo) -(2sqrt2)/sqrt(x-4) = 0#

The range for the first equation is #-oo < y < 0#

#lim_(x to 4) (2sqrt2)/sqrt(x-4) = oo#

#lim_(x to oo) (2sqrt2)/sqrt(x-4) = 0#

The range for the second equation is #0 < y < oo#