How do you find vertical, horizontal and oblique asymptotes for $f(x)= (3x^2 + 15x +18 )/( 4x^2-4)$ ?

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Answer 1 · 2016-12-04T09:57:13

Vertical: $uarr x = +-1 darr$
Horizontal: $larr y = 3/4 rarr$
Illustrative graph is inserted.

$f(x) =3/4((x+2)(x+3))/((x-1)(x+1)$

Zeros: x = 0 and 4.

AS $x to +-oo, f to 3/4$ .

As $x to +-1, y to +-oo$ .

Interestingly, the asymptote y = 3/4 cuts the graph#

at $x = (15+-sqrt323)/2= 16.48 and -1.49$ .

Yet, it is tangent at infinity.

graph{y(4x^2-4)-(3x^2+15x+18)=0 [-24.43, 24.43, -12.22, 12.22]}

How do you find vertical, horizontal and oblique asymptotes for f(x)= (3x^2 + 15x +18 )/( 4x^2-4) f(x)= (3x^2 + 15x +18 )/( 4x^2-4)?