Andre W. answered 10/07/13
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I'm assuming you mean
f(x)=(-3x³+3x²+x-7)/(2x²-6x-3)
Since this is a rational function whose numerator degree (3) is one higher than the denominator degree (2), it will have an oblique asymptote, in addition to the two vertical asymptotes you get by setting the denominator equal to zero.
To find the oblique asymptote, use long division to re-write f(x) as
f(x) = (-(43/2) x-16)/(2x²-6x-3) - (3/2)x - 3
As x→±∞, the first term goes to zero, so the oblique asymptote is given by
g(x) =-(3/2) x - 3.
It intersects the graph of f(x) when
f(x)=g(x),
which is the case when
(-(43/2) x-16)/(2x²-6x-3)=0,
which gives us x=-32/43.