Question

What is the end behavior of the equation below?

Answer 1

\[f\left(x\right)=\frac{\left(x-3\right)\left(x+4\right)}{\left(x-3\right)\left(x-8\right)}\]

Answer 2

May I help?

Answer 3

@3mar Of course :)

Answer 4

Thanks.

Answer 5

Any ideas?

Answer 6

I honestly have no idea how end behaviour works so
@3mar

Answer 7

\[f\left(x\right)=\frac{\left(x-3\right)\left(x+4\right)}{\left(x-3\right)\left(x-8\right)}=\frac{ x+4 }{ x-8 }\]

Done!

Answer 8

https://www.desmos.com/calculator/rledanzdxa
and this is the graph of the final fraction!

Answer 9

I'm a little confused though. Is end behaviour just the simplified equation or..?

Answer 10

When dealing with rational functions,
the end behavior is determined by the degree of the top in comparison to the degree of the bottom.

When they are of equal degree (as happened in this problem),
then you end up with a horizontal asymptote with value equal to the ratio of the coefficients of the leading terms.

That was kind of confusing the way I said that...
Lemme give an example:\[\large\rm \frac{3x^2-4}{2x^2+3}\]The degree of both the numerator and denominator is 2, they both have a squared x,
so we have a horizontal asymptote.
It's value is equal to the ratio of the leading coefficients,
so in this example we have a horizontal asymptote at 3/2.

Answer 11

So in your problem,\[\large\rm \frac{x+4}{x-8}\]The degrees are equal, first degree,
so again we'll end up with some sort of horizontal asymptote as end behavior.
There are no special coefficients on our x's though, right?
You can think of it like this I suppose,\[\large\rm \frac{1x+4}{1x-8}\]So our asymptote is located at 1/1.

Horizontal asymptote at y=1.
That is your end behavior.

Answer 12

AHHHH I hope that wasn't too confusing >.<
I was rambling on for a bit there..

Answer 13

@zepdrix Not confusing at all. I actually understood it pretty well. Thank you!

Answer 14

@brokenstanzas
my medal should go for zepdrix, not to me. He deserves it for good illustration and good explanation.