Expert: Paul Klarreich - 5/3/2008

Question
QUESTION: Hi, I'm studying rational functions and I'm having trouble to solve and graph this rational function:
f(x)= x^3-x^2-6x/x^2-3x+2  

I need to:
a)Factor: I already factored it but I dont have a picture to show.
b)find domain: I think its {x/x is not equal to 2, x is not equal to 1}
c) find vertical asymptotes/ holes. vertical asymptote x=0?
d) find horizontal/oblique asymptote
e) find x-y-intercept (if applies). y-int = 0?
f) graph

Can you please help me out?

ANSWER:  Questioner:   Andrey
Category:  Calculus
Private:  No
 
Subject:  Rational Functions
Question:  Hi, I'm studying rational functions and I'm having trouble to solve and graph this rational function:
f(x)= x^3-x^2-6x/x^2-3x+2  

I need to:
a)Factor: I already factored it but I dont have a picture to show.
b)find domain: I think its {x/x is not equal to 2, x is not equal to 1}
c) find vertical asymptotes/ holes. vertical asymptote x=0?
d) find horizontal/oblique asymptote
e) find x-y-intercept (if applies). y-int = 0?
f) graph

Can you please help me out?
 
a.  Looks like:
x(x - 3)(x + 2)
---------------
(x - 2)(x - 1)

b)find domain: I think its {x/x is not equal to 2, x is not equal to 1}

Right on.  AND x = 2 and x = 1  (yes, I said EQUAL) are your vertical asymptotes.

Your zeroes (x-intercepts) would be the zeroes on top,  x = 0, x = 3, x = -2.  Fortunately none matches your V.A.s, so there are no 'holes'.  You get those when you see:

(stuff)(x - a)
------------------
(more stuff)(x - a)

For the y-intercept, you just put  x = 0, and obviously get y = 0.

Asymptotes at infinity.  Go back to the UNFACTORED FORM:

x^3-x^2-6x
-----------
x^2-3x+2  

Take out the largest power of x on top and bottom:

x^3(1 - 1/x - 6/x^2)
-------------------- =
x^2(1 - 3/x + 2/x)

x(1 - 1/x - 6/x^2)
------------------
(1 - 3/x + 2/x)

Now let  x -> infinity.  All those fractions disappear.

x(1 - 0 - 0)
------------ = x.
(1 - 0 + 0)

y = x (45-degree line) is your oblique asymptote.

[If it were a constant, it would be a horizontal asymptote.]

Now you can put it all together.


---------- FOLLOW-UP ----------

QUESTION: Thank you very much for your help. I just have 2 more questions. the function is improper, right? So don't I have to do a long division? I did that and I got y=x+2 as the oblique asymptote. Another question is how is the graph going to look like between the vertical asymptotes? Is it a parabola, decreasing or increasing? I couldn't figure that out.

Answer
QUESTION: Thank you very much for your help. I just have 2 more questions. the function is improper, right?

>> Right.

So don't I have to do a long division?

I did that and I got y=x+2 as the oblique asymptote.

>> Drat!  I think I blew that one.  You are indeed correct.  Although my arithmetic is right, I only proved that

lim(x->inf) f(x)/x = 1, which is not good enough.  

Sorry.

Another question is how is the graph going to look like between the vertical asymptotes? Is it a parabola, decreasing or increasing? I couldn't figure that out.

>> My picture (see new one) shows a 'parabola-looking' piece between x=1 and x = 2, but it is not a parabola.  Here is what you must do if you don't want to spend your whole day differentiating this f(x).

To judge the behavior BETWEEN x = 1 and x = 2, determine

lim  f(x) [on the right of x = 1, OR > 1]
x->1+

lim  f(x) [on the left of x = 2, or < 2]
x->2-

Use the factored forms.  Of course both approach infinity, but you need to know whether it's  PLUS infinity oR MINUS

      (x)(x - 3)(x + 2)
lim    ------------------ =
x->1+  (x - 2)(x - 1)

      (POS)(NEG)(POS)
lim    ------------------ = POSITIVE infinity.
x->1+  (NEG)(POS)


      (POS)(NEG)(POS)
lim    ------------------ = positive infinity.
x->2-  (NEG)(POS)

Positive at both aymptotes.  So draw it as you see on the picture.

Where is that minimum point? Don't get me started on that one.