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Which statement is true about the discontinuities of the function f(x)?
f(x)=x+1/6x^2-7x-3
There are asymptotes at x=3/2 and x=-1/3
There are holes at x = 3/2 and x = -1/3
There are asymptotes at x = -3/2 and x = 1/3
There are holes at x = -3/2 and x = 1/3

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Answer:

  (a)  There are asymptotes at x=3/2 and x=-1/3

Step-by-step explanation:

The denominator zeros can be found by factoring:

  f(x) = (x +1)/((2x -3)(3x +1))

Neither of the denominator factors is cancelled by the numerator factor, so each represents a vertical asyptote, not a function hole.

The asymptotes are at the values of x where the denominator is zero:

  2x -3 = 0   ⇒   x = 3/2

  3x +1 = 0   ⇒   x = -1/3

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