How do you find the vertical, horizontal and slant asymptotes of: #f(x)= (x)/(4x^2+7x+-2)#?

1 Answer
Dec 7, 2016

Answer:

For + sign equation - Horizontal:#larr y = 0rarr#; and vertical : #uarr x =-1.39 and x = -0.34 darr#. For #-# sign- Horizontal: larr y=0rarr#; and vertical: #uarr x= -2 and x=1/4 darr#.

Explanation:

The zeros of the denominator #4x^2+7x+_2# are

#-1.39 and -0.34# for + sign #and -2 and 1/4# for - sign.

For the straight lines x = these values, #y to +-oo#.

Also, as #x to +-oo, y to 0.

graph{y(4x^2+7x+2)-x=0 [-20, 20, -10, 10]}
graph{y(4x^2+7x-2)-x=0 [-10, 10, -5, 5]}