# Calculus

posted by on .

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 - ,

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 - ,

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 - ,

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 - ,

Thank you, I see where I made my error now. X^2 over x^2 becomes 1. I cancelled them out.