## anonymous 5 years ago find the domain for y =x^2square root (x^2+5)

1. myininaya

all real numbers.

2. myininaya

that is $y=x^2*\sqrt{x^2+4}$

3. myininaya

5 not 4

4. anonymous

so how would you find the vertical asymptote ?

5. myininaya

just ask yourself where is this function not defined when where there be a negative under the square root never since x^2 is always positive so x^2+5 is always positive

6. myininaya

there are no vertical asy. the function is continuous everywhere

7. anonymous

is this since the domain is real for all values of x?

8. myininaya

yes

9. myininaya

if we had 1/(x-1) the vertical asy would be at x=1

10. anonymous

ahhhh thank you very much myininaya!

11. myininaya

if we had (x-1)/[(x-1)(x-2)] then there would be a hole at x=1 and a vertical asy at x=2

12. myininaya

since the x-1 cancel

13. anonymous

thank you ^.^