How do you find the asymptotes for #1/3(x-1)^3+2#?

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Answer 1 · 2016-12-11T14:18:34

No asymptote. This is an increasing function, making an x-intercept -0,8172 and y-intercept 5/3, nearly. See graph.

#y=1/3(x-1)^3+2 =0#, when #x = -6^(1/3)+1=-0.81712#, nearly

#y'=(x-1)^2>=0#. So, y is an increasing function, excepting at x = 1.

#y'=0, when x = 1#. Here, #y''=0 and y'''=2>0#

So, (1, 2) is a point of inflexion.

As #x to +-oo, y to +-oo#.

There is no asymptote.

graph{y-(x-1)^3/3-2=0 [-10, 10, -5, 5]}