How do you find the vertical, horizontal and slant asymptotes of: f(x)= (x)/(4x^2+7x+-2)f(x)=x4x2+7x±2?

1 Answer
Dec 7, 2016

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

Explanation:

The zeros of the denominator 4x^2+7x+_24x2+7x+2 are

-1.39 and -0.341.39and0.34 for + sign and -2 and 1/4and2and14 for - sign.

For the straight lines x = these values, y to +-ooy±.

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]}