Question 1210370
<pre>
I corrected my former answer.

{{{drawing(400,400,-6,6,-6,6,
graph(400,400,-6,6,-6,6,(x+1)/(sqrt(x^2 - 5))),
green(line(-7,-1,-1,-1)),
green(line(-7,1,7,1), line(-sqrt(5),-7,-sqrt(5),7), line(sqrt(5),-7,sqrt(5),7)


)
 )}}} 


{{{y=(x+1^"")/sqrt(x^2-5)}}}

There are vertical asymptotes at {{{x="" +- sqrt(5)}}},
and there are no points between those two vertical asymptotes.

We know that there is a vertical asymptote at {{{x=sqrt(5)}}} so y goes to infinity on the right.  

Also since the limit of y as x approaches +infinity is 1, there is a horizontal
asymptote y = 1 and y is always above that horizontal asymptote when
{{{x>sqrt(5)}}}

Also since the limit of y as x approaches -infinity is -1, there is a horizontal
asymptote y = -1 and y is always above that horizontal asymptote when
{{{x<sqrt(5)}}}



{{{y*sqrt(x^2-5)=x+1}}}
{{{y^2(x^2-5)=(x+1)^2}}}
{{{y^2x^2-5y^2 = x^2+2x+1}}}
{{{y^2x^2 - x^2- 5y^2-2x-1=0}}}
{{{x^2(y^2-1)-2x-(5y^2+1) = 0}}}
{{{Discriminant = B^2-4AC = (-2)^2-4(y^2-1)(-(5y^2+1)^"")}}}
{{{Discriminant = 4-4(y^2-1)(-5y^2-1)}}}
Factor the negative out of the last factor:
{{{Discriminant = 4+4(y^2-1)(5y^2+1)}}}
{{{Discriminant = 4+4(5y^4+y^2-5y^2-1)=4+4(5y^4-4y^2-1)=4+20y^4-16y^2-4)}}}
{{{Discriminant = 20y^4-16y^2}}}
The discriminant is not negative:
{{{20y^4-16y^2>=0}}}
{{{4y^2(5y^2-4)>=0}}}
{{{y^2(5y^2-4)>=0}}}
Since {{{y^2>=0}}} is always true, we must have
{{{5y^2-4>=0}}}
{{{5y^2>=4}}}
{{{y^2>=4/5}}}

Since y is always negative left of its left asymptote

{{{y<-sqrt(4/5)}}}

However y is always above its horizontal asymptote y = 1 to the right of its
vertical asymptote 

Since the denominator is always positive, y is always negative when x<-1 

So on the left of the vertical asympote. {{{y<=-sqrt(4/5)}}} 

However on the right of its horizontal asymptote {{{x=sqrt(5)}}} y is always
above its horizontal asymptote y = 1, so y > 1 when {{{x>sqrt(5)}}}

So the range of the function is

{{{matrix(1,3,

matrix(1,5,"(",-infinity,",",-sqrt(4/5),"]"),
 U, 
matrix(1,5,"(",1,",",infinity,")"))}}}

or you can rationalize the denominator:

{{{matrix(1,3,

matrix(1,5,"(",-infinity,",",-(2sqrt(5))/5,")"),
 U, 
matrix(1,5,"(",1,",",infinity,")"))}}}

Edwin</pre>