Question

which statement is true about the discontinuities of the function ( f(x) )?
( f(x)=\frac{x + 1}{6 x^{2}+7 x - 3} )
there are holes at ( x=-\frac{3}{2} ) and ( x=\frac{1}{3} ).
there are holes at ( x=\frac{3}{2} ) and ( x=-\frac{1}{3} ).
there are asymptotes at ( x=-\frac{3}{2} ) and ( x=\frac{1}{3} ).
there are asymptotes at ( x=\frac{3}{2} ) and ( x=-\frac{1}{3} ).

Explanation:

Step1: Find the denominator of the function

For a rational function \(y = \frac{f(x)}{g(x)}\), the function is discontinuous when \(g(x)=0\).
Given \(f(x)=\frac{x + 1}{6x^{2}-7x - 3}\), we first factor the denominator.
We use the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\), where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\).
For \(6x^{2}-7x - 3\), \(a = 6\), \(b=-7\), \(c=-3\).
\(x=\frac{7\pm\sqrt{(-7)^{2}-4\times6\times(-3)}}{2\times6}=\frac{7\pm\sqrt{49 + 72}}{12}=\frac{7\pm\sqrt{121}}{12}=\frac{7\pm11}{12}\)
\(x_1=\frac{7 + 11}{12}=\frac{3}{2}\), \(x_2=\frac{7-11}{12}=-\frac{1}{3}\)

Step2: Analyze the discontinuity

Since the numerator \(x + 1\) and the denominator \(6x^{2}-7x - 3=(2x - 3)(3x+1)\) have no common factors, the function \(y=\frac{x + 1}{6x^{2}-7x - 3}\) has vertical asymptotes at the values of \(x\) that make the denominator equal to zero.

Answer:

There are asymptotes at \(x=\frac{3}{2}\) and \(x =-\frac{1}{3}\)