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:

Step1: Factor the denominator

To find discontinuities, factor \(6x^2 - 7x - 3\). We need two numbers \(a\) and \(b\) such that \(a\times b=6\times(-3)= - 18\) and \(a + b=-7\). The numbers are \(-9\) and \(2\).

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Step2: Find the values of \(x\) that make denominator zero

Set \((3x + 1)(2x - 3)=0\).

• For \(3x+1 = 0\), we have \(3x=-1\), so \(x =-\frac{1}{3}\).
• For \(2x - 3=0\), we have \(2x=3\), so \(x=\frac{3}{2}\).

Step3: Check for holes or asymptotes

A hole occurs when a factor cancels out from numerator and denominator. The numerator is \(x + 1\), and it does not have factors \((3x + 1)\) or \((2x - 3)\). So, there are no common factors, which means these values of \(x\) (where denominator is zero) are vertical asymptotes, not holes.

Answer:

There are asymptotes at \(x=\frac{3}{2}\) and \(x =-\frac{1}{3}\) (the first option).