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

1 Answer
Jan 8, 2017

Answer:

Horizontal : #larr y = 0 rarr#
Vertical : #uarr x = -2 darr#.

Explanation:

Let #y = (2x^2-1)/(3x^3-2x+1)#

There is no quotient in this division. So, existence of slant asymptote

is ruled out..

The y-intercept ( x = 0 ) is #-1#.

x-intercepts ( y = 0 ) are #x=+-1/sqrt2#.

Cross multiplying,

#y (3x^3-2x+1) = 2x^2-1#

If one factor in LHS is 0, the other has to be infinite, when the

product #in (0, oo)#.

Now, y = 0 gives horizontal asymptote.

The other factor has a factor (x +1 ) and its other factor has complex

zeros.

So, x + 2 = 0 gives the vertical asymptote.

graph{ y(y (3x^3-2x+1) - 2x^2+1)=0 [-8, 8 -3.26, 3.276]} #

See the Socratic graph.