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]}

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