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

Aug 8, 2017

Please consider the x term. Is there any real number that one can substitute into that term which would make the function be undefined. No. Therefore, x is any real number:

$x \in \mathbb{R}$

The y term is similar to the x term, except that is has a negative sign, therefore, the same is true for the y term:

$x , y \in \mathbb{R}$

This constitutes the domain.

$f \left(3 , - 1\right) = {\left(3 - 3\right)}^{2} / 4 - {\left(\left(- 1\right) + 1\right)}^{2} / 16$
$f \left(3 , - 1\right) = 0$
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 $x \to \pm \infty$. 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 $y \to \pm \infty$. This adds the negative real numbers to the range:
$f \left(x , y\right) \in \mathbb{R}$