Question

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

Answer 1

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.