Find the horizontal or oblique asymptote of f(x) = negative 3 x squared plus 7 x plus 1, all over x minus 2
Finding the horizontal and oblique asymptote of
f(x)= (-3x^2+7x+1)/(x-2)
Solution:
For Horizontal Asymptote:
Line
y=L is a horizontal asymptote of the function y=f(x), if either limx→∞f(x)=Llimx→∞f(x)=L or limx→−∞f(x)=Llimx→−∞f(x)=L, and LL is finite.
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.
