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.

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

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.

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 !

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

To determine if the function f(x) = (x^2 - 3x + 2)/(1 + x^2) has a horizontal asymptote, we need to analyze the behavior of the function as x approaches positive or negative infinity.

First, you correctly divided all terms in the numerator and denominator by x^2. However, it seems like there is an error in simplifying the expression.

After dividing by x^2, the simplified expression should be:

(-3/x) + (2/x^2) / (1/x^2)

To simplify this further, we combine the fractions in the numerator:

(-3/x + 2/x^2) / (1/x^2)

The next step is to simplify the numerator by finding a common denominator:

(-3x + 2) / x^2 / (1/x^2)

Now, let's rewrite the expression as a division of fractions:

(-3x + 2) / x^2 * x^2 / 1

By canceling out the common factor x^2 in the numerator and denominator, we get:

-3x + 2

Now, let's analyze the behavior of this expression as x approaches positive or negative infinity. As x approaches infinity, the term -3x dominates the expression because the x^2 term becomes insignificant compared to x. Therefore, the function f(x) does not approach a constant value as x approaches infinity or negative infinity.

Hence, the function f(x) = (x^2 - 3x + 2)/(1 + x^2) does not have a horizontal asymptote.