Question

which statement is true about the discontinuities of the function \\(f(x)\\)?

\\f(x) = \frac{x+1}{6x^2-7x-3}\\

there are asymptotes at \\(x = \frac{3}{2}\\) and \\(x = -\frac{1}{3}\\).
there are holes at \\(x = \frac{3}{2}\\) and \\(x = -\frac{1}{3}\\).
there are asymptotes at \\(x = -\frac{3}{2}\\) and \\(x = \frac{1}{3}\\).
there are holes at \\(x = -\frac{3}{2}\\) and \\(x = \frac{1}{3}\\).

Explanation:

Factor the denominator

Using the Polynomial Factoring knowledge point

$$ 6x^2 - 7x - 3 = (2x - 3)(3x + 1) $$

Rewrite the function

Using the Rational Functions knowledge point

$$ f(x) = \frac{x + 1}{(2x - 3)(3x + 1)} $$

Identify the discontinuities

Using the Vertical Asymptotes and Holes in Rational Graphs knowledge points

$$
LATEXBLOCK0
$$

Since neither factor in the denominator cancels with the numerator \(x + 1\), both values correspond to vertical asymptotes.

Answer:

• (A) There are asymptotes at \(x = \frac{3}{2}\) and \(x = -\frac{1}{3}\). (Correct answer)
• (B) There are holes at \(x = \frac{3}{2}\) and \(x = -\frac{1}{3}\).
• (C) There are asymptotes at \(x = -\frac{3}{2}\) and \(x = \frac{1}{3}\).
• (D) There are holes at \(x = -\frac{3}{2}\) and \(x = \frac{1}{3}\).