<span>Find the horizontal or oblique asymptote of f(x) = negative 3 x squared plus 7 x plus 1, all over x minus 2
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Finding the horizontal and oblique asymptote of
f(x)= (-3x^2+7x+1)/(x-2)
Solution:
For Horizontal Asymptote:
Line <span>y=L</span><span> is a horizontal asymptote of the function </span><span>y=f<span>(x)</span></span><span>, if either </span><span><span><span>lim<span>x→∞</span></span>f<span>(x)</span>=L</span><span><span>lim<span>x→∞</span></span>f<span>(x)</span>=L</span></span><span> or </span><span><span><span>lim<span>x→−∞</span></span>f<span>(x)</span>=L</span><span><span>lim<span>x→−∞</span></span>f<span>(x)</span>=L</span></span><span>, and </span><span>LL</span><span> is finite.
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Calulate limits:
<span><span>lim<span>x→∞</span></span><span>(<span><span>1<span>x−2</span></span><span>(<span>−3<span>x2</span>+7x+1</span>)</span></span>)</span>=−∞</span>
<span><span>lim<span>x→−∞</span></span><span>(<span><span>1<span>x−2</span></span><span>(<span>−3<span>x2</span>+7x+1</span>)</span></span>)</span>=∞</span>
Thus, there are no horizontal asymptotes.
For Oblique Asymptote:
Do polynomial long division <span><span>(<span>−3<span>x2</span>+7x+1</span>)/(x-2)</span>=−3x+1+<span>3/(<span>x−2)</span></span></span>
The rational term approaches 0 as the variable approaches infinity.
Thus, the slant asymptote is <span>y=−3x+1y=−3x+1</span>.
