Respuesta :
Using the discontinuity concept, it is found that:
There are asymptotes at [tex]x = \frac{3}{2}[/tex] and [tex]x = --\frac{1}{3}[/tex]
------------------------------------
A function is discontinuous at a point outside it's domain, and at these points in the graph there will be asymptotes.
------------------------------------
The given function is:
[tex]f(x) = \frac{x + 1}{6x^2 - 7x - 3}[/tex]
It is a fraction, then the denominator cannot be zero, and the asymptotes will be at the values of x for which:
[tex]6x^2 - 7x - 3 = 0[/tex]
------------------------------------
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
------------------------------------
For this equation, [tex]a = 6, b = -7, c = -3[/tex]. Thus:
[tex]\Delta = (-7)^{2} - 4(6)(-3) = 121[/tex]
[tex]x_{1} = \frac{-(-7) + \sqrt{121}}{2(6)} = \frac{18}{12} = \frac{3}{2}[/tex]
[tex]x_{2} = \frac{-(-7) - \sqrt{11}}{2(6)} = -\frac{4}{12} = -\frac{1}{3}[/tex]
Thus, there are asymptotes at [tex]x = \frac{3}{2}[/tex] and [tex]x = --\frac{1}{3}[/tex]
The graph of the function is given at the end of the answer.
A similar problem is given at https://brainly.com/question/23214539
