Question

find the domain for
y =x^2square root (x^2+5)

Answer 1

all real numbers.

Answer 2

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

Answer 3

5 not 4

Answer 4

so how would you find the vertical asymptote ?

Answer 5

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

Answer 6

there are no vertical asy. the function is continuous everywhere

Answer 7

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

Answer 8

yes

Answer 9

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

Answer 10

ahhhh thank you very much myininaya!

Answer 11

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

Answer 12

since the x-1 cancel

Answer 13

thank you ^.^