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Question 215432: Another one I can't figure out.
y=x^2/(x^2-16) state the domain and range
I figured out the domain (I think):
x^2-16=0
(x-4)(x+4)=0
x=-4 or x=+4
The domain is "all x not equal to -4 or 4". Is that correct?
And I can't find the range, can anybody help?
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You are spot on, the domain is 
The range is the set of values that the function can assume for all possible values of the independent variable in the domain.
Let's look at values of  in the interval  . The square of any value in this interval is going to be smaller than 16, so the value of the function on this interval will be negative or zero. Hence, on this interval, the maximum value of the function is zero, and there is no minimum value of the function because you can continue to choose values of closer and closer to either 4 or -4 making the function get closer and closer to  .
Values of the independent variable either slightly greater than 4 or slightly less than -4 will result in a very large positive value for the function. Again, you can make the function as large as you want by selecting a value as close as you want to either -4 or 4.
As the absolute value of the independent variable gets very large, that is as you approach either or  , the constant term in the denominator has less and less effect on the overall value of the quotient. For example, if  you have  , a little less than 3. But if  , you have  a number much closer to 1. In fact, 1 is the limiting number. For any value  or  , the function will always be 
Taking it all together, you can see that the range excludes anything in the interval  , hence the range is defined as
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! y=x^2/(x^2-16) state the domain and range
I figured out the domain (I think):
x^2-16=0
(x-4)(x+4)=0
x=-4 or x=+4
The domain is "all x not equal to -4 or 4". Is that correct?
And I can't find the range, can anybody help?
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The domain is correct.
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Range?
Notice that the horizontal asymptote is x^2/x^2 = 1
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So the graph would never cross the line y = 1.
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Since the numerator is larger than the denominator,
y is above y=1 for x^2>16 (i.e. x>4 or x<-4)
and y is below y=0 for x^2 < 16 (i.e. -4< x <4
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Looks like the Range is All Real Numbers except 0< y <1
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Cheers,
Stan H.
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