How do you find the vertical, horizontal or slant asymptotes for $f(x) = (3x^2 + 4)/(x+1)$ ?

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Answer 1 · 2017-06-29T12:03:24

Vertical asymptote at $x= -1$ , Slant asymptote: $y=3x-3$ ,
Horizontal asymptote: Absent

$f(x) =(3x^2+4)/(x+1)$
Vertical asymptote : Denominator is $0 :. x+1 = 0 or x = -1$

Vertical asymptote at $x= -1$

Degree i.e maximum power of $x$ in numerator is $1$ more than denominator, so there is no horizontal asymptote.

Slant asymptote: Degree i.e maximum power of $x$ in numerator is $1$ more than denominator , so there is slant asymptote.

By long division of (f(x) we get dividend as $y=3x-3$ and remainder $7 :. y = 3x-3$ is the slant asymptote.

graph{(3x^2+4)/(x+1) [-80, 80, -40, 40]} [Ans]

How do you find the vertical, horizontal or slant asymptotes for f(x) = (3x^2 + 4)/(x+1)f(x) = (3x^2 + 4)/(x+1)?