Please help, I'm being dumb
2 answers:
You might be interested in
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
