Consider the graph of \(\frac{x^3x^2+x}{6x^29x}.\) 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.
\(y\ =\ \frac{x^3x^2+x}{6x^29x}\)
To find the number of holes, let's factor the numerator and denominator like this: 

\(y\ =\ \frac{x(x^2x+1)}{x(6x9)}\)  
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