Expert: Paul Klarreich - 12/5/2007

Question
I have no clue how to do this problem and i've been working on it for quit a while now. graph for f(x) = x-3 / x^2 + x-2 by finding the intercepts, relative extrema, inflection points, behavior of where f is undefined, and behavior at positive and negative infinity
Thanks jess

Answer
Questioner:   jess
Category:  Calculus
Private:  No
 
Subject:  calculus homework question
Question:  I have no clue how to do this problem and i've been working on it for quit a while now. graph for f(x) = x-3 / x^2 + x-2 by finding the intercepts, relative extrema, inflection points, behavior of where f is undefined, and behavior at positive and negative infinity
Thanks jess
...................................
Hi, Jess,

Actually, what you have is a set of detailed instructions for drawing (I like to call it sketching) the graph.  A SKETCH is a picture that shows the features of the graph:

A. the intercepts.

-- these are the points where the graph touches or crosses an axis.  

A1 Find a y-intercept by setting x = 0 and determining f(0) = y. Note the point and make sure your graph shows it.

A2. Find x-intercept(s) by solving f(x) = 0.  


B Find asymptotes

B1. IF any x makes a denominator zero, that is (probably) a vertical asymptote.  Draw  vertical, dotted line.

B2. If lim (x -> inf) = a const, that is a horizontal asympote.  Draw a horizontal dotted line at the right.

B3. Same for  lim (x -> - inf)

C. Get f'(x).  Find where it is zero.  That's a 'stationary' point, probably a max/min.  Find where f' is positive (graph rises) and where it is negative (graph falls).

D. Get f''.  Likewise, get inflection points.

Mark all these features, then use your basic detective skills to put an entire graph together.

[YOU THOUGHT CALCULUS WOULD BE EASY?]

BTW, your example as written, was:
           3  
f(x) = x - --- + x - 2
          x^2  

Did you really mean that?  If not, and you meant:
         x - 3
f(x) = ---------------
      x^2 + x - 2

then write that, in the future.  You can write:

f(x) = (x-3)/(x^2 + x-2)

Don't like using parentheses?  Time for a career change.

Anyway, here is how it goes:


         x - 3
f(x) = ---------------
      x^2 + x - 2

         -3
A. f(0) = --- = +1.5.   Mark  (0,1.5), make sure your graph shows it.
         -2

Set f(x) = 0.  Just the top is enough.  x = 3.  Mark (3,0), make sure your graph shows it.


B. Set  x^2 + x - 2 = 0
   (x + 2)(x - 1) = 0
x = -2, x = 1 are vertical asymptotes.
Draw them, dotted.

         x -3
lim  --------------- = 0. So  y = 0 is an asymptote.
      x^2 + x - 2
x-> +-inf

           (x^2 + x - 2)(1) - (x - 3)(2x + 1)  
C. f'(x) =  -----------------------------------
                  ()^2


         x^2 + x - 2 - (2x^2 - 5x  - 3 )
f'(x) =  -----------------------------------
                  ()^2

         x^2 + x - 2 - 2x^2 + 5x  + 3
f'(x) =  ------------------------------
                  ()^2

          - x^2 + 6x  + 1
f'(x) =  ------------------------------
                  ()^2

Set - x^2 + 6x  + 1 = 0

x^2  - 6x  + 9       = 1 + 9

(x - 3)^2 = 10

x = 3 +- sqrt(10), about  x = 6 and x = - 0.2

You can probably do the rest, including using a graphing calculator to get a general idea.  But make sure to mark the features.