Question

Find the horizontal asymptotes of:

Answer 1

\[y=\sqrt{x ^{2}+1}/3x\]

Answer 2

First simplify/solve the nominator to x+1. Then continue on the equation to x+1/3x.
Then look at the highest degrees of the x on both top and bottom of the equation and that would be x and 3x. The rule says that if they are equal, the horizontal asyptote is the line so it's 1/3

Answer 3

@kimchriss I agree with your answer of 1/3 but \[\sqrt{x^2+1}\] does not reduce to X+1

Answer 4

so the answer is 1/3

Answer 5

\[y=\frac{ \sqrt{x^2+1} }{ 3x}=\frac{ \sqrt{x^2(1+\frac{ 1 }{ x^2 })} }{ 3x }=\frac{ x \sqrt{1+\frac{ 1 }{ x^2 }} }{ x(3) }=\frac{ 1 }{ 3 }\]

Answer 6

\[\pm \frac{ 1 }{ 3 }\] because of the square root, hope that helps!

Answer 7

it does thank you!!!

Answer 8

np(: