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

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Answer 1 · 2017-01-08T16:06:11

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

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

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.

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