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

Question

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

1 Answer

Impact of this question

1296 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-23T15:27:03

The vertical asymptotes are $x = \frac{1}{2}$ $x=1/2$ and $x = - 2$ $x=-2$

The horizontal asymptote is $y = \frac{1}{2}$ $y=1/2$

Explanation:

We facorise $y = \frac{(x + 1) (x - 1)}{(2 x - 1) (x + 2)}$ $y=((x+1)(x-1))/((2x-1)(x+2))$
As we cannot divide by $0$ $0$
So for $x = \frac{1}{2}$ $x=1/2$ and $x = - 2$ $x=-2$ we have vertical asymptotes
To look for horizontal asymptotes, we rewrite
$y$ $y$ as $y = \frac{1 - \frac{1}{x^{2}}}{2 + \frac{3}{x} - \frac{2}{x^{2}}}$ $y=(1-1/x^2)/(2+3/x-2/x^2)$

the limit $y = \frac{1 - 0}{2 + 0 + 0} = \frac{1}{2}$ $y=(1-0)/(2+0+0)=1/2$
$y \to - \infty$ $y->-oo$

the limit $y = \frac{1 - 0}{2 + 0 + 0} = \frac{1}{2}$ $y=(1-0)/(2+0+0)=1/2$
$y \to + \infty$ $y->+oo$
So $y = \frac{1}{2}$ $y=1/2$ is a horizontal asymptote

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the vertical, horizontal or slant asymptotes for y=(x^2-1)/(2x^2 + 3x-2)y=x2−12x2+3x−2y=(x^2-1)/(2x^2 + 3x-2)?

1 Answer

Explanation:

Related questions

Impact of this question

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