# Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of the rational function?

x^2-5x-6/2x^2-x-21 I also need the domain of f.

f(x) = (x²-5x-6)/(2x²-x-21)

= (x-6)(x+1)/[(2x-7)(x+3)]

Removable discontinuities do not exist : numerator and denominator do not share a factor

Zeros exist at x = -1 and x = 6

Vertical asymptotes exist at x = -3, and x = 7/2 : denominator has zeros at x = -3 and 7/2

Horizontal asymptote exists at y = 1/2 : lim(x→±∞) f(x) = 1/2

Oblique asymptotes do not exist : numerator and denominator are the same order

Domain of f(x) is all x except x = -3 and x = 7/2 : x in (-∞, -3)U(-3, 7/2)U(7/2, ∞)

2x^2-x-21 = 2x^2-6x+7x-21 = 2x(x-3)+7(x-3) = (x-3)(2x+7)

(x^2-5x-6) /(2x^2-x-21)

= (x^2-5x-6) /((x-3)(2x+7))

The vertical asymptote is x=3 and x = -7/2

The horizontal asymptote:

As x approaches infinity, the numerator and the denominator being of the same degree, approaches 1/2

The coefficient of x^2 in the numerator is 1 and in the denominator it's 2

y= 1/2 is the horizontal asymptote

x^2-5x-6/2x^2-x-21

= [(x - 6)(x + 1)]/[(2x - 7)(x + 3)]

Vertical asymptotes: x = 7/2, and - 3

(1 - 5/x - 6/x^2)/(2 - 1/x - 21/x^2)

As x ---- > - ∞, f(x) ------> 1/2

As x ------> .∞ f(x) -----> 1/2

Horizontal asymptote: y = 1/2

No oblique asymptote

Domain = R - { 7/3, -3}

2x²-x-21 ≠ 0 for all ℝ, so there are no vertical asymptotes.