Question

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

( f(x)=\frac{x - 5}{3 x^{2}-17 x - 28} )

there are holes at ( x = 7 ) and ( x=-\frac{4}{3} ).

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 ( (-7,0) ) and ( x=\frac{4}{3} ).

Explanation:

Step1: Factor the denominator

Factor \(3x^{2}-17x - 28\).
We need to find two numbers \(a\) and \(b\) such that \(a\times b=3\times(- 28)=-84\) and \(a + b=-17\). The numbers are \(-21\) and \(4\).

$$3x^{2}-17x - 28=3x^{2}-21x + 4x-28=3x(x - 7)+4(x - 7)=(3x + 4)(x - 7)$$

So, \(f(x)=\frac{x - 5}{(3x + 4)(x - 7)}\)

Step2: Analyze discontinuities

A rational function \(y=\frac{N(x)}{D(x)}\) has vertical asymptotes at the values of \(x\) for which \(D(x)=0\) and \(N(x)
eq0\).
Set \(D(x)=(3x + 4)(x - 7)=0\).
Solving \(3x+4 = 0\) gives \(x=-\frac{4}{3}\), and solving \(x - 7=0\) gives \(x = 7\).
Since the numerator \(N(x)=x - 5\) is not zero when \(x=-\frac{4}{3}\) (\(N(-\frac{4}{3})=-\frac{4}{3}-5=-\frac{4 + 15}{3}=-\frac{19}{3}
eq0\)) and \(N(7)=7 - 5=2
eq0\)

Answer:

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