1. domain

y=(x^2+40x-500)/500

i thought the domain would be all real numbers since there's no radical, no chance for 0s in the denominator, but the book says the domain is 500<x<750. how did they get that?

2. Originally Posted by ihateregistering
y=(x^2+40x-500)/500

i thought the domain would be all real numbers since there's no radical, no chance for 0s in the denominator, but the book says the domain is 500<x<750. how did they get that?
500y = x^2+40x-500

500y / (x^2+40x-500) = 1

so x^2 + 40x <> 500

3. are you sure the book asks for the domain? if it does, then the book is wrong and the domain is all real numbers

4. I think that it's the range that we have to find not the domain.

5. Originally Posted by roshanhero
I think that it's the range that we have to find not the domain.

if they're asking for the range then there is no upper bound on what the y values can be since the limit as x goes to infinity is infinity

this thing is clearly a parabola which is concave up, i am not sure why the book gives that answer... are you sure you're looking at the right problem

6. Originally Posted by Wilmer
500y = x^2+40x-500

500y / (x^2+40x-500) = 1

so x^2 + 40x <> 500
Sorry this is wrong. The given function is a typical parabola. The domain is all real numbers.

Furthermore, if you complete the square you get $\displaystyle y = \frac{1}{500} (x + 20)^2 - \frac{9}{5}$ and the clearly the range has nothng to do with the given answer either. As has already been remarked, IF the posted question is accurate then the book's answer is wrong.

7. Not the first typo I've found. Cliffs Notes no bueno. Thanks for confirming that I know how to find a domain. I was worried!