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Consider the graph of \(\frac{x^3-x^2+x}{6x^2-9x}.\) Let a be the number of holes in the graph, b be the number of vertical asympotes, c be the number of horizontal asymptotes, and d be the number of oblique asymptotes. Find a+2b+3c+4d.

 Jul 28, 2019
 #1
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\(y\ =\ \frac{x^3-x^2+x}{6x^2-9x}\)

 

To find the number of holes, let's factor the numerator and denominator like this:

 

 

\(y\ =\ \frac{x(x^2-x+1)}{x(6x-9)}\)  
There is only  1  common factor between the numerator and denominator so there is only  1  hole.

 

 

a = 1  
   
The vertical asymptotes occur when the denominator after common factors are removed equals zero.

 

 

6x - 9  =  0  
6x  =  9

 

 

x  =  3/2  
There is only  1  vertical asymptote.

 

 

b = 1  
   
Since the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes.

 

 

c = 0  
   
The degree of the numerator is exactly one greater than the degree of the denominator, so there is an oblique asymptote.

 

 

d = 1  

 

Check: https://www.desmos.com/calculator/vrya3tghg2

 

a + 2b + 3c + 4d   =   1 + 2(1) + 3(0) + 4(1)   =   1 + 2 + 4   =   7

 Jul 28, 2019

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