Posted by **Greg** on Friday, September 14, 2012 at 12:19pm.

Check my work please?

Determine if f(x)= (x^2-3x+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 x-axis, 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

(x-1)(x-2) , so clearly we have 2 x-intercepts

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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