HA crossover and holes

[Aethez]

I stumbled across this function
f(x)=(x^2+8x+16)/(x^2+x-12)

Factoring this:
=[(x+4)^2]/[(x+4)(x-3)]
=(x+4)/(x-3)

where x cannot equal -4 and 3

Since I cancelled x+4, there is a hole at x =-4
Therefore there is a vertical asymptote at x=3

Solving for the y value at x=-4 (hole), y=0

There is a horizontal asymptote at x=1

This is where I was confused:
to check for crossovers on the HA, I set 1=f(x)
Here I got x=-4 which is the x value of the hole.

Basically, my question is: How come there is a hole at (-4,0) and then a possible crossover at (-4,1)?

 

[Aethez]

It seems that there is no solution (no crossover) when I set the factored form of the function = 1, however I was taught to always work with the original function for crossovers?

 

[Dr.Peterson]

There is a horizontal asymptote at x=1

You mean y = 1, of course.

This is where I was confused:
to check for crossovers on the HA, I set 1=f(x)
Here I got x=-4 which is the x value of the hole.

Basically, my question is: How come there is a hole at (-4,0) and then a possible crossover at (-4,1)?

Please show your work to get x = -4. I think you made a mistake there. (When things don't make sense, that's a good possibility to consider first.)

It seems that there is no solution (no crossover) when I set the factored form of the function = 1, however I was taught to always work with the original function for crossovers?

That's correct. There is no "crossover" where the graph of the function crosses its HA:


It makes no difference in this case which form you use.

 

[Steven G]

Please don't say Solving for the y value at x=-4 (hole), y=0
When x=-4, there is NO y value, after all, there is a hole there. So y is not 0 or any other number!

 

[Steven G]

It seems that there is no solution (no crossover) when I set the factored form of the function = 1, however I was taught to always work with the original function for crossovers?

What you were taught is not wrong, but you can use the reduced function to set equal to -1. If you happen to compute that x=-4, then you should reject that value.

Using the reduced function is always going to be easier to use.