Question

How do you find vertical, horizontal and oblique asymptotes for `(2x^2 - x - 38) / (x^2 - 4)`?

Answer 1

`"vertical asymptotes at " x=+-2`
`"horizontal asymptote at " y=2`

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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 " x^2-4=0rArr(x-2)(x+2)=0`

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

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" ( a constant)"`

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

`f(x)=((2x^2)/x^2-x/x^2-38/x^2)/(x^2/x^2-4/x^2)=(2-1/x-38/x^2)/(1-4/x^2)`

as `xto+-oo,f(x)to(2-0-0)/(1-0)`

`rArry=2" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 2) hence there are no oblique asymptotes.
graph{(2x^2-x-38)/(x^2-4) [-20, 20, -10, 10]}