Find the horizontal or oblique asymptote of f(x) = negative 3 x squared plus 7 x plus 1, all over x minus 2

Calulate limits:

limx→∞(1x−2(−3x2+7x+1))=−∞

limx→−∞(1x−2(−3x2+7x+1))=∞

Thus, there are no horizontal asymptotes.

For Oblique Asymptote:

Do polynomial long division (−3x2+7x+1)/(x-2)=−3x+1+3/(x−2)

The rational term approaches 0 as the variable approaches infinity.

Thus, the slant asymptote is y=−3x+1y=−3x+1.

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step-by-step explanation:

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i got 10.28 but that rounds up to 10.3