find the slant y = (2x^2 Answer step by step.

You need to remember the formula of slant asymptote such
that:

`f(x) = ax + b`

`a =
lim_(x-gt+-oo)(f(x))/x`

`b = lim_(x-gt+-oo)(f(x) -
ax)`

You need to evaluate a such
that:

`lim_(x-gt+-oo)(f(x))/x =
lim_(x-gt+oo)(2x^2-x-8)/(x(2x+3))`

`lim_(x-gt+-oo)(2x^2-x-8)/(2x^2+3x)`

You
need to factor out `x^2`  to numerator and denominator such
that:

`lim_(x-gt+-oo)(x^2(2-1/x-8/x^2))/(x^2(2+3/x))`

Reducing
by `x^2`  yields:

`lim_(x-gt+-oo)(2-1/x-8/x^2)/(2+3/x) =
(2-0-0)/(2+0) = 2/2 = 1`

`a =
1`

Since the value of a exists, then you may evaluate b
such that:

`b = lim_(x-gt+-oo)(2x^2 - x - 8 - x) =
oo`

Hence,  since b fails to exist, then the
function has no slant asymptotes to `+-oo.`