Question

which statement is true about the discontinuities of the function $f(x)$?
$f(x) = \frac{x - 5}{3x^2 - 17x - 28}$
there are asymptotes at $x = -7$ and $x = \frac{4}{3}$.
there are asymptotes at $x = 7$ and $x = -\frac{4}{3}$.
there are holes at $x = 7$ and $x = -\frac{4}{3}$.
there are holes at $(-7, 0)$ and $x = \frac{4}{3}$.

Explanation:

Step1: Factor Denominator

Factor \(3x^2 - 17x - 28\). We need two numbers \(a,b\) such that \(a \cdot b = 3 \cdot (-28)= -84\) and \(a + b = -17\). The numbers are \(-21\) and \(4\). So, \(3x^2 - 21x + 4x - 28 = 3x(x - 7) + 4(x - 7) = (3x + 4)(x - 7)\). Thus, \(f(x)=\frac{x - 5}{(3x + 4)(x - 7)}\).

Step2: Analyze Discontinuities

For rational functions, vertical asymptotes occur where the denominator is zero and the numerator is not zero (no common factors). Holes occur where numerator and denominator have common factors (but here numerator \(x - 5\) and denominator \((3x + 4)(x - 7)\) have no common factors). So, set denominator zero: \((3x + 4)(x - 7)=0\) gives \(x = 7\) (from \(x - 7 = 0\)) and \(x = -\frac{4}{3}\) (from \(3x + 4 = 0\)). These are vertical asymptotes (since no common factors with numerator).

Answer:

There are asymptotes at \(x = 7\) and \(x = -\frac{4}{3}\) (the second option).