question about long division, please help!! (1 Viewer)

[Dm31]

how do u find the horizontal asymptote when
x(x-1)
-------
x-2 ???????

i kno the answer is meant to be like x, and i'm meant to do long division but i completely forgot!!

i also know that curve sketching is meant to be the easiest part of 4u, yet i'm having problems with it...

should i be extremely worried?

 

[dawso]

long divison?? i dunno wats goin on there but wen i do these, i just think of the horizontal assumptote as wen x=a bill-i-on(say that in the nice dr evil voice, adds charm)

so this would be (a bill-i-on) (a billion minus one)
. ----------------------------------------
. abillion minus two


=abillion
=x

therefore, as u said, it is at y=x

thats prob no help cause u got that but i dont know wat ya mean by ya long division, sorray

-dawso

 

[dawso]

in regards to whether u shud be worried - no, this topic just comes with practice, doin questions and even knowing ur calculus and shit better all helps ya, u get it eventually

 

[martin]

Yeah what you're thinking of is a way of writing it as a polynomial plus a ratio of polynomials with the top one being of smaller degree than the bottom. This way isn't long division but you might like it better. Basically you try and write the top so you can split it up and then cancel with the bottom.

x*(x-1)/(x-2)
=(x^2-x)/(x-2)
=(x^2-2x+x)/(x-2)
=x*(x-2)/(x-2) + x/(x-2)
=x + (x-2+2)/(x-2)
=x + (x-2)/(x-2) + 2/(x-2)
= x + 1 +2/(x-2)

so as x -> infinity, 2/(x-2) -> 0 so the horizontal asymptote is y = x + 1.

I can't think of a good way to explain long division, I'll try and write it up later.

 

[Slidey]

how do u find the horizontal asymptote when
x(x-1)
-------
x-2 ???????

i kno the answer is meant to be like x, and i'm meant to do long division but i completely forgot!!

i also know that curve sketching is meant to be the easiest part of 4u, yet i'm having problems with it...

should i be extremely worried?

Expand top to get:
(x^2-x)/(x-2)

Complete the square on the top: (x^2-x -3x + 4 +3x -4)/(3x-2)
=[(x-2)^2 + 3x-4]/(x-2)
=(x-2) + (3x-4)/(x-2)
Again: (3x-4)/(x-2)=(3x-4-2x+2-2+2x)/(x-2)=(x-2 -2+2x)
=1-2(x-1)/(x-2)
And again:
2(x-1)/(x-2)=2[1+1/(x-2)]=2+2/(x-2)
So: (x-2)+1+2+2/(x-2)
So
(x+1)+2/(x-2)

You've got an oblique asymptote at y=x+1.

The reason I did it that way was two-fold: One, I wanted to show you how to do it that way, and two: You can't really do long multiplication or synthetic on the computer.

EDIT: Martin beat me to it and he did so with a more elegant solution.

 

[martin]

Alright I've written up a long division method. If you remember standard long division it isn't too bad. First open the picture then read the description

write x-2)x^2-x (with a line over the top)

look at the highest powers in the divisor (x - 2) and the thing being divided (x^2 -x). We see that x goes into x^2 x times so write x above. Then multiply x by x-2 and put this below and subtract. (x^2-x) - (x^2-2x) = x so write this below a line. Now x goes into x 1 time so write this above. Now x-1*(x-2) = 2 so put this below a line. x-2 doesn't divide 2 so we have a remainder 2.

Now just like 7/3 is 2 remainder 1 so 7/3 = 2 1/3

then (x^2-x) divided by (x-2) is x+1 remainder 2

so (x^2-x)/(x-2) = x + 1 + 2/(x-2)

 

[Dm31]

ThanX

Hi again,

thanks all for your replys, i get the division now and thou im still tryn to figure out sum of the other stuff there >.<

Thanks again!!!

P.S. man u people are damn smart!!!