Horizontal (slant) Aymptote

[blitzen]

I have the equation (x^3 - 3x^2) / (x^2 - 4)

Based on this, I will have a slant asymptote. After doing the long division, I got:
x-3 with a remainder of -4x + 12.

That cannot be right, can it? If it is, what do I do with the remainder?

 

[mmm4444bot]

After doing the long division, I got:

x - 3 with a remainder of -4x + 12

Your quotient is correct, but your remainder has some sign issues.


what do I do with the remainder?

How could I know? You did not post the instructions for this exercise. I mean, what did they ask you to do?

The quotient represents a linear function. Its graph is the slant asymptote because the ratio which is the remainder over the divisor tends toward zero (globally).



MY EDITS: Corrected my lapse in logic.

 

[blitzen]

Oh wow, it is -4x - 12 ! Oops, haha.

So my slant asymptote is -4x - 12 ? And that is all there is to the answer?

He just asked for horizontal and vertical asymptotes. I already found the vertical.

 

[mmm4444bot]

Oh wow, it is -4x - 12

I get 4x - 12.


that is all there is to the answer?

Yes, for the question, "What is the slant asymptote".

"The slant asymptote is y = x - 3."

MY EDIT: Fixed my continued lapse in logic.

 

[blitzen]

You are correct, a negative times a negative definitely gives me a positive....

Too many careless mistakes. Thank you!

 

[mmm4444bot]

Speaking of carelessness, I fatally switched the asymptote. (Thank you, wjm11, for correcting me.)

The remainder is written over the divisor, and, since the divisor (2nd-degree polynomial) grows much faster than the remainder (linear polynomial) for large absolute-values of x, that ratio goes toward zero, leaving x - 3 as the global behavior.

Hence, the slant asymptote is y = x - 3.

I will fix my prior posts. Please excuse my carelessness. If you need further explanation to understand the above, let me know.

Cheers ~ Mark

 

[daon]

While the division method always works, this problem is much simpler. You may factor the numerator to get:

\(\displaystyle \frac{x^2}{x^2-4}(x-3)\)

As x blows up, you can see the first factor approaches 1. So for large x, our value is close to x-3.

This will always work if you can draw out a factor of same degree as the divisor.