Question 1200106
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The denominator is {{{sqrt(x^2-1)}}}, so the radicand {{{x^2-1}}} must be non-negative.<br>
Furthermore, since that square root is in the denominator,the radicand can't be zero.<br>
{{{x^2-1=(x+1)(x-1)}}}, so the function is undefined on [-1,1].<br>
For x values a tiny bit greater than 1, the denominator is close to 0, and the numerator is positive, so the function value is large positive.  That makes x=1 a vertical asymptote.<br>
Similarly, for x values a tiny bit less than -1, the denominator is close to 0, and here the numerator is negative, so the function value is large negative.  So x=-1 is also a vertical asymptote.<br>
For either large positive or large negative values of x, {{{sqrt(x^2-1)}}} is very close to {{{sqrt(x^2)=abs(x)}}}, so the function is very nearly equal to {{{x/abs(x)}}}.<br>
For large positive values of x, {{{x/abs(x)=1}}}, so y=1 is a horizontal asymptote for x>1.<br>
For large negative values of x, {{{x/abs(x)=-1}}}, so y=-1 is a horizontal asymptote for x<-1.<br>
ANSWERS:
vertical asymptotes at x=-1 and x=1;
horizontal asymptotes at y=1 (for positive x values) and y=-1 (for negative x values).<br>
A graph of the function and the two horizontal asymptotes....<br>
{{{graph(400,400,-3,3,-3,3,x/sqrt(x^2-1),-1,1)}}}<br>