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

##### 1 Answer
Dec 11, 2016

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

#### Explanation:

$y = \frac{1}{3} {\left(x - 1\right)}^{3} + 2 = 0$, when $x = - {6}^{\frac{1}{3}} + 1 = - 0.81712$, nearly

$y ' = {\left(x - 1\right)}^{2} \ge 0$. So, y is an increasing function, excepting at x = 1.

$y ' = 0 , w h e n x = 1$. Here, $y ' ' = 0 \mathmr{and} y ' ' ' = 2 > 0$

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

As $x \to \pm \infty , y \to \pm \infty$.

There is no asymptote.

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