Question

How do you find the vertical, horizontal or slant asymptotes for `(4+x^4)/(x^2-x^4)`?

Answer 1

Vertical asymptotes at `x=-1`, `x=0` and `x=1` and horizontal asymptote at `y=-1`.

Explanation:

We have vertical asymptotes for `f(x)` at `x=a`, when `lim_(x->a)f(x)=+-oo`. If `f(x)` is an algebraic function, which happens for values of `x`, for which denominator is `0`.

Here in denominator, we have `x^2-x^4=x^2(1-x^2)` and this is equal to zero when `x=0,1` or `-1`.

Hence, we have vertical asymptotes at `x=-1`, `x=0` and `x=1`.

We have horizontal asymptote when `lim_(x->+-oo)f(x)=k`, a constant. This hapens in algebraic functions, when degree of numerator is equal to that of denominator, which is the case here.

Here `lim_(x->+-oo)(4+x^4)/(x^2-x^4)`

= `lim_(x->+-oo)(4/x^4+1)/(1/x^2-1)`

= `1/(-1)=-1`

Hence we have a horizontal asymptote `y=-1`

Slant asymptote is there when degree of numerator is just one more than that of denominator. Here we do not have a slant asymptote.

graph{(4+x^4)/(x^2-x^4) [-10, 10, -10, 25]}

As an example of slant asymptote consider `(4+x^5)/(x^2-x^4)`

`lim_(x->+-oo)(4+x^5)/(x^2-x^4)`

= `lim_(x->+-oo)(4/x^4+x)/(1/x^2-1)`

= `x/(-1)=-x` and slant asymptote is `y=-x` or `x+y=0`

graph{(4+x^5)/(x^2-x^4) [-10, 10, -10, 25]}

Graphs not drawn to scale.