Identify the oblique asymptote of f(x) = quantity x plus 4 over quantity 3 x squared plus 5 x minus 2.

                     3x² + 5x - 2 = 0
                3x² + 6x - x - 2 = 0
3x(x) + 3x(2) - 1(x) - 1(2) = 0
          3x(x + 2) - 1(x + 2) = 0
                  (3x - 1)(x + 2) = 0
3x - 1 = 0       or       x + 2 = 0
    + 1 + 1                     - 2  - 2 
     3x = 1          or          x = -2
      3     3                       1     1
       x = ¹/₃         or         x = -2

      f(x) = 3x² + 5x - 2
     f(¹/₃) = 3(¹/₃)² + 5(¹/₃) - 2
     f(¹/₃) = 3(¹/₉) + 1²/₃ - 2
     f(¹/₃) = ¹/₃ - ¹/₃
     f(¹/₃) = 0
(x, f(x)) = (¹/₃, 0)

      f(x) = 3x² + 5x - 2
     f(-2) = 3(-2)² + 5(-2) - 2
     f(-2) = 3(4) - 10 - 2
     f(-2) = 12 - 12
     f(-2) = 0
(x, f(x)) = (-2, 0)

    Vertical Asymptotes: ¹/₃ or -2
Horizontal Asymptotes: 0
      Oblique Asymptote: No Asymptotes

The answer to your question is b