How do you find the domain and range of #f(x,y) = (x-3)^2 /4 - (y+1)^2 /16#?

Question

Eli Assouline · Accepted Answer

Please take a look at the x term. Is there any real number that can be used in place of that term to render the function undefined? If not, then x can be any real number:

#x in RR#

With the exception of its negative sign, the y term and the x term are similar; hence, the following holds true for the y term:

#x,y in RR#

The domain is made up of this.

For the range, please take into account:

#f(3,-1) = (3-3)^2 /4 - ((-1)+1)^2 /16#

#f(3,-1)= 0#

Can we do something that makes the function start from zero and approach positive infinity? Yes. We can hold y constant at -1 and let #x to +-oo#. Therefore, the range is at least all positive real numbers.

Can we do something that makes the function start from zero and approach negative infinity? Yes. We can hold x constant at 3 and let #y to +-oo#. This adds the negative real numbers to the range:

#f(x,y) in RR#

Kennedy Adams · Answer

undefined The domain of the function is all real numbers for both x and y. The range of the function is y such that y is less than or equal to 0.