Question

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

Answer 1

you just have to plug in the answer and see which one equals up to the question

Answer 2

Answer:

There are asymptotes at:

             $x=\dfrac{-1}{3}\ and\ x=\dfrac{3}{2}$

Step-by-step explanation:

In a rational function the holes are the points where both numerator as well as denominator of the given rational function is zero.

Also, if the degree of the numerator is smaller than the degree of the denominator then the rational function has horizontal asymptote at y=0

and the vertical asymptotes are the values of x where the denominator of the rational function is equal to zero.

We are given a rational function f(x) by:

      $f(x)=\dfrac{x+1}{6x^2-7x-3}$

This could also be represented by:

$\dfrac{x+1}{6x^2-7x-3}=\dfrac{x+1}{6x^2-9x+2x-3}\\\\i.e.\\\\\dfrac{x+1}{6x^2-7x-3}=\dfrac{x+1}{3x(2x-3)+1(2x-3)}\\\\i.e.\\\\\dfrac{x+1}{6x^2-7x-3}=\dfrac{x+1}{(3x+1)(2x-3)}$

Hence, the numerator is zero when :

$x=-1$

and the denominator is zero when,

$x=\dfrac{-1}{3}\ and\ x=\dfrac{3}{2}$

Since, there is no such x such that the numerator and denominator both are zero.

Hence, the rational function does not has holes.

Also, the vertical asymptotes are at:

$x=\dfrac{-1}{3}\ and\ x=\dfrac{3}{2}$

( Since, the denominator is zero at that point)