what is the asymptotes for y=3x-1/√9x^2-1

Question

anonymous · Answer

You will have a vertical asymptote at the place where the denominator is equal to zero, hence the equation is not defined.

anonymous · Answer

for the remaining asymptotes, you can see how this equations acts as the limit of x approaches infinity.

anonymous · Answer

but the answer is y=+1 and -1,x=+1/3 and -1/3

anonymous · Answer

$$ \sqrt{9x^2-1}=0 \
9x^2=1
\x= \pm \frac{1}{3} $$

anonymous · Answer

for horizontal asymptote?

anonymous · Answer

Sorry I had to disappear for a moment.

anonymous · Answer

You know how horizontal asymptotes can be found?

$$ \lim_{x \to \infty} \frac{3x-1}{\sqrt{9x^2-1}} $$

anonymous · Answer

the annoying part is that you only have a radical in the denominator, but you can easily help yourself to get rid of this expression. 

$$\pm (3x-1) = (\sqrt{3x-1})^2 $$
See where this is going?

anonymous · Answer

$$ \sqrt{\frac{9x^2-6x+1}{9x^2-1}} $$ Equal grade between numerator and denominator, hence the horizontal limit is:

$$ y= \pm 1 $$