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

Answer 1

Julia Adams

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#

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Answer 2

Kennedy Adams

To find the domain and range of the equation ( (y^2)(x-4) = 8 ), consider the restrictions on ( x ) and ( y ). The domain is all real numbers except ( x = 4 ), and the range is all real numbers except ( y = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.