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posted by  kennethp on 7/19/2009 8:06:52 PM  |  status: Live  |  Earned Karma: 25

domain, asymptotes, pre-cal

Course Textbook Chapter Problem Needs by
Precalculus Precalculus (3rd) by Blitzer N/A N/A N/A
Question Details:
Given the rational function  f(x)= (3x^3+5) / (x+1)^2 (x-2)

A)Find the domain of the function
b)Find the horizontal asymptote for the graph of the function
c) find the 2 vertical asymptotes for the graph of the function

Please be clear as i would like to understand this problem.... kenny, thanks ahead of time
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Sage
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posted by ayusuf on 7/19/2009 8:17:57 PM  |  status: Live
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Response Details:
a.) The domain is the set of all real numbers except -1 and 2.
b.) I don't think there are any horizontal asymtotes.
c.) The vertical asymtotes are x = -1 and x =2. because they are the zeroes for denominator.

Please Rate and PM me if you have further questions. Thanks.
Sage
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posted by kinshuk (MNK) on 7/19/2009 8:23:42 PM  |  status: Live
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Response Details:
Given the rational function  f(x)= (3x^3+5) / (x+1)^2 (x-2)
 

A)
   domain of the function:-
-----------------------------
f(x)= (3x^3+5) / (x+1)^2 (x-2)  .
clearly f(x) will be undefined if the denominator is zero as you cant divide anything by zero.
so, f(x) will be undefined if 
                                       (x+1)^2 * (x-2) = 0 
                                
either (x+1)^2 = 0       or         (x - 2) = 0 
    => x + 1 = 0            or         x = 2
    => x = -1                 or         x = 2
therefore domain of definition of the function f(x) :-
                                                      -α < x < α , but x  -1 and x 2      .....................(Answer)
 
plz do rate the answer if you liked it
Oracle
Karma Points: 14,954
posted by Grace on 7/20/2009 1:12:19 AM  |  status: Live
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Response Details:
Don't forget to rate!



A) The domain is all x-values except where there are holes or vertical asymptotes.
There are no holes since nothing cancels across the numerator/denominator line.
Vertical asymptotes occur where the denominator equals 0:
(x + 1)2(x-2) = 0
x = -1, x = 2
So domain is (-∞, -1) U (-1, 2) U (2, ∞).

B) You need to multiply everything out and then look at the largest degree of x in the top and bottom to determine if there are horizontal asymptotes.
=
Since the leading term in the top and bottom is the same the horizontal asymptote is their coefficients so HA is y = 3.

C) see work in part A.
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