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]}