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

Answer 1

Elias Ackerman

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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Answer 2

Anna Allen

To find the asymptotes of the function ( \frac{1}{3}(x-1)^3 + 2), first, identify if there are any vertical asymptotes by checking for values of (x) that make the denominator zero. Since there are no denominators in the expression, there are no vertical asymptotes.

Next, determine if there are any horizontal asymptotes. For polynomial functions, the degree of the numerator and denominator determines the behavior as (x) approaches positive or negative infinity. In this case, the degree of the numerator is 3, and there's no denominator, so there are no horizontal asymptotes.

Therefore, the function does not have any asymptotes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.