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

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0that is \[y=x^2*\sqrt{x^2+4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so how would you find the vertical asymptote ?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0just 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

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0there are no vertical asy. the function is continuous everywhere

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this since the domain is real for all values of x?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if we had 1/(x1) the vertical asy would be at x=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhhh thank you very much myininaya!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if we had (x1)/[(x1)(x2)] then there would be a hole at x=1 and a vertical asy at x=2
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