Question

which statement is true about the discontinuities of the function f(x)?
$f(x)=\frac{x - 5}{3x^2 - 17x - 28}$

• 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 denominator

Factor $3x^2 -17x -28$:
$3x^2 -17x -28 = (3x+4)(x-7)$

Step2: Check common factors

Numerator is $x-5$, which has no common factors with $(3x+4)(x-7)$.

Step3: Find discontinuities

Set denominator to 0:
$(3x+4)(x-7)=0$
Solve for $x$:
$3x+4=0 \implies x=-\frac{4}{3}$
$x-7=0 \implies x=7$
Since no common factors, these are vertical asymptotes.

Answer:

There are asymptotes at $x = 7$ and $x=-\frac{4}{3}$