Question

How do you find the asymptotes for `y= (x^2-1)/(2x^2 + 3x-2)`?

Answer 1

Vertical asymptotes occur when the doniminator is zero

i.e. when `2x^2 + 3x - 2 = 0`

or, (2x - 1)(x + 2) = 0

Hence, when `x = 1/2 and x = -2`

Horizontal asymptotes occur as the limiting value of the function y

Dividing by `x^2` we have:

y = `(x^2 - 1)/(2x^2 + 3x - 2)` => `(1 - 1/x^2)/(2 + 3/x - 2/x^2)`

As x --> ± ∞, y --> 1/2

Hence, the line `y = 1/2` is a horizontal asymptote

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Answer 2

`"vertical asymptotes at " x=-2" and "x=1/2`

`"horizontal asymptote at "y=1/2`

Explanation:

The denominator of y cannot be zero as this would make y `color(blue)"undefined".`Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

`"solve " 2x^2+3x-2=0rArr(2x-1)(x+2)=0`

`rArrx=-2" and " x=1/2" are the asymptotes"`

`"Horizontal asymptotes occur as"`

`lim_(xto+-oo),ytoc" (a constant)"`

divide terms on numerator/denominator by the highest power of x, that is `x^2`

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

as `xto+-oo,yto(1-0)/(2+0-0)`

`rArry=1/2" is the asymptote"`
graph{(x^2-1)/(2x^2+3x-2) [-10, 10, -5, 5]}