+0

0
73
1

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
+8724
+4

$$y\ =\ \frac{x^3-x^2+x}{6x^2-9x}$$

 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

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

Jul 28, 2019