To determine the true statement about the discontinuities of the function [tex]\( f(x) = \frac{x+1}{6x^2 - 7x - 3} \)[/tex], one must find the values of [tex]\( x \)[/tex] where the function is not defined, which occur when the denominator equals zero.
1. Identify the denominator's equation:
[tex]\[
6x^2 - 7x - 3 = 0
\][/tex]
2. Find the roots of the denominator:
Solving the quadratic equation [tex]\( 6x^2 - 7x - 3 = 0 \)[/tex], we get two solutions. The roots of this quadratic equation are:
[tex]\[
x = \frac{3}{2} \quad \text{and} \quad x = -\frac{1}{3}
\][/tex]
3. Classify the discontinuities:
Since the numerator [tex]\( x + 1 \)[/tex] does not cancel with any factors in the denominator [tex]\( 6x^2 - 7x - 3 \)[/tex], these points are vertical asymptotes. Thus, there are vertical asymptotes (not holes) at these values of [tex]\( x \)[/tex].
Given the possible statements:
- There are asymptotes at [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( x = -\frac{1}{3} \)[/tex]:
This is true because these are the points where the denominator is zero and the function is not defined, indicating vertical asymptotes.
- There are holes at [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( x = -\frac{4}{3} \)[/tex]:
This is false because the points of discontinuity are vertical asymptotes and not holes. Additionally, [tex]\( x = -\frac{4}{3} \)[/tex] is incorrect as it is not a root of the denominator.
- There are asymptotes at [tex]\( x = -\frac{3}{2} \)[/tex] and [tex]\( x = \frac{1}{3} \)[/tex]:
This is false because these values are not the roots of the denominator, hence not the points of discontinuities.
- There are holes at [tex]\( x = -\frac{3}{2} \)[/tex] and [tex]\( x = \frac{1}{3} \)[/tex]:
This is also false for the same reason as above, these points are not where the denominator equals zero.
Therefore, the correct answer is that:
There are asymptotes at [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( x = -\frac{1}{3} \)[/tex].
