Question

which statement is true about the discontinuities of the function $f(x)$?
$f(x) = \frac{x + 1}{6x^2 - 7x - 3}$

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 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

The denominator is \(6x^2 - 7x - 3\). We factor it:

$$
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$$

So the function becomes \(f(x)=\frac{x + 1}{(3x + 1)(2x - 3)}\).

Step2: Analyze discontinuities

For a rational function, holes occur when there is a common factor in the numerator and denominator. Here, the numerator \(x + 1\) and the factored denominator \((3x + 1)(2x - 3)\) have no common factors. So, there are no holes. Vertical asymptotes occur where the denominator is zero (and numerator is non - zero).
Set the denominator equal to zero:
\((3x + 1)(2x - 3)=0\)
Using the zero - product property:
\(3x+1 = 0\) gives \(x=-\frac{1}{3}\) and \(2x - 3=0\) gives \(x=\frac{3}{2}\)
Wait, let's re - check the factoring. Wait, maybe I made a mistake in factoring. Let's use the quadratic formula for \(ax^{2}+bx + c = 0\), \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(6x^{2}-7x - 3\), where \(a = 6\), \(b=-7\), \(c=-3\)
\(x=\frac{7\pm\sqrt{(-7)^{2}-4\times6\times(-3)}}{2\times6}=\frac{7\pm\sqrt{49 + 72}}{12}=\frac{7\pm\sqrt{121}}{12}=\frac{7\pm11}{12}\)
So \(x=\frac{7 + 11}{12}=\frac{18}{12}=\frac{3}{2}\) and \(x=\frac{7-11}{12}=\frac{-4}{12}=-\frac{1}{3}\)
Wait, the original denominator factoring was wrong. Let's do it again:
\(6x^{2}-7x - 3\)
We need two numbers that multiply to \(6\times(-3)=-18\) and add up to \(-7\). The numbers are \(-9\) and \(2\)
So \(6x^{2}-9x+2x - 3 = 3x(2x - 3)+1(2x - 3)=(3x + 1)(2x - 3)\) is correct. So the roots of the denominator are \(x =-\frac{1}{3}\) (from \(3x + 1 = 0\)) and \(x=\frac{3}{2}\) (from \(2x - 3=0\))
Since the numerator \(x + 1\) is not zero at \(x=-\frac{1}{3}\) (when \(x =-\frac{1}{3}\), \(x + 1=-\frac{1}{3}+1=\frac{2}{3}
eq0\)) and at \(x=\frac{3}{2}\) (when \(x=\frac{3}{2}\), \(x + 1=\frac{3}{2}+1=\frac{5}{2}
eq0\)), these are vertical asymptotes.

Looking at the options, the third option says "There are asymptotes at \(x=\frac{3}{2}\) and \(x =-\frac{1}{3}\)".

Answer:

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