What happens if the Vertical asymptote has the same X co-ordinate as the hole?

[I.Need.Help]

The function is (X^4+x^3-10x^2+8x)/(-x+4). I got a hole at the x value of 2, but the VA I got when setting the denominator to zero was -2 and 2. Would I exclude the VA of 2 because that is the X value of the hole?

 

[tkhunny]

1) How did you get 2 and -2?


2) Note: [math]y = \dfrac{x^{2}-4}{x-2}[/math] is EXACTLY the same as [math]y = x+2[/math] EXCEPT at x = 2. Why? Does this help?

 

[I.Need.Help]

1) How did you get 2 and -2?


2) Note: [math]y = \dfrac{x^{2}-4}{x-2}[/math] is EXACTLY the same as [math]y = x+2[/math] EXCEPT at x = 2. Why? Does this help?

Sorry I made a mistake in my question, I meant to write -x^2+4 as the denominator

 

[Dr.Peterson]

The function is (X^4+x^3-10x^2+8x)/(-x+4). I got a hole at the x value of 2, but the VA I got when setting the denominator to zero was -2 and 2. Would I exclude the VA of 2 because that is the X value of the hole?

If it's a hole, then it's not an asymptote. Your check for an asymptote should exclude holes; I do so by working from the canceled (hole-less) function rather than from the original.

But the function you wrote has 0 in the denominator at x = 4, not at x = 2 and x = -2. Did you type it wrong?

In any case, please show more of your work, so we can point out where it needs to be modified.

 

[I.Need.Help]

If it's a hole, then it's not an asymptote. Your check for an asymptote should exclude holes; I do so by working from the canceled (hole-less) function rather than from the original.

But the function you wrote has 0 in the denominator at x = 4, not at x = 2 and x = -2. Did you type it wrong?

In any case, please show more of your work, so we can point out where it needs to be modified.

Yeah I meant to write -x^2+4

 

[I.Need.Help]

If it's a hole, then it's not an asymptote. Your check for an asymptote should exclude holes; I do so by working from the canceled (hole-less) function rather than from the original.

But the function you wrote has 0 in the denominator at x = 4, not at x = 2 and x = -2. Did you type it wrong?

In any case, please show more of your work, so we can point out where it needs to be modified.

Just to clarify, once there is a hole I should start working with the function that it had been crossed off from right? Instead of continuing from the original?

 

[Dr.Peterson]

Just to clarify, once there is a hole I should start working with the function that it had been crossed off from right? Instead of continuing from the original?

Right.

Sometimes you will find that there is still an uncanceled factor in the denominator (e.g. if there had been an (x-2) in the numerator and (x-2)^2 in the denominator, so you are left with (x-2) in the denominator) so that you do in fact have an asymptote there (not a hole), so that in a sense an asymptote takes priority over a hole. So I really should have said, "if there's an asymptote, it's not a hole". If the factor is entirely eliminated from the denominator, as usually happens, then it is really a hole.