Discover the Asymptotes of (x-1)^2: A Comprehensive Guide

[frosty8688]

1.Find the asymptotes of the following equation.



2. $\left.x/(x-1)^2\right.$



3. I know that the asymptote as x approaches ±∞ is ±∞. I am wondering when (x-1)$^{2}$ approaches ±∞ is also ±∞

 

[Mark44]

1.Find the asymptotes of the following equation.



2. $\left.x/(x-1)^2\right.$

To be an equation, there needs to be an = somewhere.

3. I know that the asymptote as x approaches ±∞ is ±∞.

?
As x gets large or very negative, the denominator of x/(x - 1)² is much larger than the numerator.

I am wondering when (x-1)$^{2}$ approaches ±∞ is also ±∞

Now I think I know what you're trying to say. You seem to be using "asymptote" in place of "limit." The two words are not synonyms, and don't mean the same thing.

As x approaches ∞, both x and (x - 1)² approach ∞. As x approaches -∞, x approaches -∞, but (x - 1)² approaches +∞. Another way to say this is
$$\lim_{x \to -\infty} (x - 1)^2 = \infty $$

The thing is, the denominator gets large much more quickly than the numerator. One way to approach this problem is to factor x² out of each term in the denominator, and factor x² out of the numerator.

 

[frosty8688]

So there would be no asymptotes.

 

[Mark44]

So there would be no asymptotes.

Do you understand what an asymptote is? How would you define it in your own words?

 

[frosty8688]

A vertical asymptote exists when x = a and a is determined by the values at which x is 0 and exists when f(x) = plus or minus infinity and a horizontal asymptote exists when x approaches infinity the number becomes larger and larger and if there is a denominator, the values when the x is in the denominator approach 0. So, there would be a horizontal asymptote at the x-axis. No vertical asymptotes, since the function approaches 0 as x goes to plus or minus infinity.

 

[eumyang]

No vertical asymptotes

This is wrong. Take the denominator and set equal to 0 to find the vertical asymptote(s).

 

[frosty8688]

Vertical asymptote at x = 1.

 

[HallsofIvy]

Vertical asymptote at x = 1.

Yes, that is correct. Do you understand why eumyang said

Take the denominator and set equal to 0 to find the vertical asymptote(s).

?