Question

f(x)=(x^2 -12x+32)/(x^12 -8x^11) Vertical Asymptote(s) x=?

Answer 1

No clue where to even go with this one, the numerator easily factors to (x-4)(x-8) but the denominator is such a mess i'm lost.

Answer 2

For the denominator:\[x^{12}-8x^{11}=x^{11}\left(x-8\right)\]
Vertical asymptotes occurs for values of \(x\) that make the denominator = 0.

Answer 3

Oh i didn't even think of that. Well wouldn't (x-8) cancel out with the top. So you would be left with (x-4)/(x^11)?

Answer 4

So there are no asymptotes, since the denominator can't be 0 unless x =0?

Answer 5

Well its not x=0 and 8, and its not no asymptotes. I don't see what else it could be,really confused now.

Answer 6

So you have
\[f(x)=\frac{x^2-12x+32}{x^{12}-8x^{11}}=\frac{(x-4)(x-8)}{x^{11}(x-8)}=\frac{x-4}{x^{11}}\]
When \(x=0\), the denominator is also zero, and so the function is undefined. In other words, \(f(0)\) does not exist. So, a vertical asymptote for the function occurs at \(x=0\).

As for when \(x=8\): keep in mind that the function is undefined when \(x=8\).
\[f(8)=\frac{(8-4)(8-8)}{8^{11}(8-8)}=\text{does not exist}\]
However, no vertical asymptote occurs. Why? Because when you canceled out the powers of \(x-8\) in both numerator and denominator, you \(removed\) it from the function. \(f(8)\) doesn't exist; instead, you have a hole in the function. This is called a "removable discontinuity."

Answer 7

Right that makes sense, but what would i put as the answer then? It says No asymptote is the wrong answer, must be something.

Answer 8

There's one asymptote at \(x=0\).

Answer 9

How?

Answer 10

I thought you said my explanation made sense? I pointed out why in that wall of text.

Answer 11

I misunderstood, i just reread and it makes sense now. I thought you were saying there were no asymptotes. Thanks

Answer 12

You're welcome!

Answer 13

There is a slant asymptote at y=x-8.