f(x) = (2x - 3) / [(2x - 3)(4x + 1)]
There are two x-values at which this rational function is undefined, x = 3/2 and x = - 1/4, the zeros of the denominator. These create two, different, discontinuities in the graph:
Because the factor (2x - 3) appears in the numerator AND denominator, this will create a hole (removable discontinuity) in the graph, at x = 3/2. To determine the y-value of this hole, cancel the (2x - 3)'s and then plug x = 3/2 into what remains: 1 / [4(3/2) + 1] = 1/7. So there is a removable discontinuity at (3/2 , 1/7).
x = - 1/4 is a vertical asymptote for the graph, a fundamental discontinuity. The left-hand branch of the graph starts in QIII by "hugging" its horizontal asymptote at y = 0, then decreases without bound, disappearing into - ∞ as x → - 1/4 from the left (ie x < - 1/4). The right-hand branch re-emerges at + ∞ for x → - 1/4 from the right (ie x > - 1/4), decreases as it passes through its y-int, (0 , 1), heading toward its horizontal asymptote at y = 0.
The horizontal asymptote mentioned above exists because the degree of the numerator is less than the degree of the denominator, so HA at y = 0. Btw, this is NOT a discontinuity; instead, it is a description of the end behavior of this rational function (as x → - ∞ and as x → ∞).
