Posted by Greg on Friday, September 14, 2012 at 12:19pm.
Check my work please?
Determine if f(x)= (x^23x+2)/(1+x^2) has a horizontal asymptotes. If so, give the equation.
So I took x^2 and divided everything by it.
(x^2/x^2)(3x/x^2)+(2/x^2)
/
(1/x^2)+(x^2/x^2)
Cancelled out some x's and have
(3/x)+(2/x^2)
/
(1/x^2)
Now as x approaches infinity they all end up becoming 0?
Making the answer No there are no Horizontal asymptotes.

Calculus  Reiny, Friday, September 14, 2012 at 1:55pm
picking it up from your :
So I took x^2 and divided everything by it.
(x^2/x^2)(3x/x^2)+(2/x^2)/(1/x^2)+(x^2/x^2)
= (1  3/x + 2/x^2 )/(1/x^2 + 1)
for Horizontal asymtotes we look at what happens to the function as x >∞
As x > ∞ , the terms 3/x , 2/x^2 , and 1/x^2 all go to zero
so we are left with
f(x) = (1  0 + 0)/(0 + 1) = 1
so the horizontal axis is
y = 1
(you appear to have dropped the value of 1 from both the top and the bottom )
btw: the entire graph lies above the line y=1

Calculus  ahem  Steve, Friday, September 14, 2012 at 2:22pm
since the denominator is always positive, yet the numerator is negative for 1<x<2, the graph has to dip below the xaxis, which means it cannot always lie above y=1.

good catch Steve  Calculus  Reiny, Friday, September 14, 2012 at 9:41pm
of course !!
I should have noticed that the numerator factors to
(x1)(x2) , so clearly we have 2 xintercepts
which of course lie below y = 1
How silly of me to make that statement at the end, and it wasn't even part of the question !

Calculus  Greg, Monday, September 17, 2012 at 4:19pm
Thank you, I see where I made my error now. X^2 over x^2 becomes 1. I cancelled them out.
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