Question

Does the function x^3-4x^2+2x-6/x-3 have a slant asymptote? if so, find an equation of the slant asymptote. if not, explain.

Answer 1

No. We do not have a slant asymptote as the degree of numerator is more than that of denominator by two. A slant asymptote would be there only if this difference is exactly one.

Explanation:

No. We do not have a slant asymptote as the degree of numerator is more than that of denominator by two. A slant asymptote would be there only if this difference is exactly one.

Let us consider `lim_(x->oo)(x^3-4x^2+2x-6)/(x-3)`

= `lim_(x->oo)(x^2-4x+2-6/x)/(1-3/x)`

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

= `x^2-4x+2`

Had it been `(x^3-4x^2+2x-6)/(x-3)`, we would have got

`lim_(x->oo)(-4x^2+2x-6)/(x-3)`

= `lim_(x->oo)(-4x+2-6/x)/(1-3/x)`

= `(-4x+2)/1`

= `-4x+2`

and slant asymptote would have been `y=-4x+2`