anonymous
  • anonymous
find the domain for y =x^2square root (x^2+5)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
all real numbers.
myininaya
  • myininaya
that is \[y=x^2*\sqrt{x^2+4}\]
myininaya
  • myininaya
5 not 4

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anonymous
  • anonymous
so how would you find the vertical asymptote ?
myininaya
  • 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
myininaya
  • myininaya
there are no vertical asy. the function is continuous everywhere
anonymous
  • anonymous
is this since the domain is real for all values of x?
myininaya
  • myininaya
yes
myininaya
  • myininaya
if we had 1/(x-1) the vertical asy would be at x=1
anonymous
  • anonymous
ahhhh thank you very much myininaya!
myininaya
  • 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
myininaya
  • myininaya
since the x-1 cancel
anonymous
  • anonymous
thank you ^.^

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