# Asamatopes -> 3U graphes.... (1 Viewer)

#### Enigma17

##### New Member
i was wondering if anyone else had problems trying to graph these things?

the ones where it looks like this:

......(x - 1)( x - 2)
y=..------------------..................(looks pretty crap)
......(x*2- 9)(x + 4)..................( x*2 = x squared)

well anyway, i know that x cannot equal 3 and -3, and -4... so there are asymatopes... but i dont know what to really do after that...
my teacher did some crap like "when it approaches from 3 on the + side, where does it go? + infinity or - infinity" or some crap like that..........

anyone got any tips on how to do them....

#### Lazarus

##### Retired
If you don't want to use the normal method (looking at the behaviour of the curve as it approaches the asymptotes), there's another, slightly time-consuming method you can use:

y =....(x - 1)(x - 2)
.......(x^2 - 9)(x + 4)

y =.....____x^2 - 3x + 2____
..........x^3 + 4x^2 - 9x - 36

Graph these curves:
f(x) = x^2 - 3x + 2
g(x) = x^3 + 4x^2 - 9x - 36

At every value of x, divide f(x) by g(x). If you draw the first two curves accurately, it's usually fairly simple to see how the final curve will look once you've divided the two curves at a few key points.

#### Enigma17

##### New Member
what do u mean dividing it?

do u find these hard?

#### Lazarus

##### Retired
For every value of x, divide the corresponding y-coordinate of f(x) by the y-coordinate of g(x). The result of that division will give you the y-coordinate of the final curve for that value of x.

I don't find them to be that hard, but I've already done the HSC.

#### Lugia

##### Member
you know you can always try numbers... like if you want to know what happens when x approaches 3 on the + side, you put into the equation a number like 3.001 and work out the y value on your calculator. In this case you get 47.67.... so u can sorta tell that the y is getting bigger. so u can say that it approaches + infinity.

And say if u want to see what happens to the curve as x approaches infinity, sub in x = 10000.

good luck!

#### p00_p00

##### Member
yeah testing no.s is always the best and aint as time consumin as ppl think!

#### -=«MÄLÅÇhïtÊ»=-

##### Gender: MALE!!!
to find the horizontal assymptotes, expand ur equation of graph, divide top and bottom by the x with highest degree (power)

the limit when x-->infitinity is you horizontal assymptote

And then by looking at degree of your equation and the sign of x, u'll have a general clue of wat curve looks like

finally if u still not sure, u can sub in soem easy values like x-0, x=1 to check

#### Enigma17

##### New Member
well, from what you all have told me i still haven't any clue how to do them...
i think im just going to take the advice of subbing in the numbers.. its the only way i'll be able to do it...

#### Lazarus

##### Retired
Most classes spend time revising topics after the trial exams - ask your teacher to go over it.

#### Nelly

##### Member
Na man, that takes ages. Just to what Laz said. Find limits, as +-x-> oo. Also, find where to crosses x,y axis. Then after doing heeps of em, you get used to what their like

#### Minai

##### Alumni
wat about those asympotes where y=x? is that 2U or 3U?

That's 3u.

#### -=«MÄLÅÇhïtÊ»=-

##### Gender: MALE!!!
u dont really find assymptotes in 4u
or find things like turning points etc like in 3u.

#### Enigma17

##### New Member
well looks like im getting the hang on them.. so its pretty cool... but, i cant find how to find the lines y=x and y=1 and so on.. asymatopes.... how do u work them out algebracially?

like
..........x^2
y =_______
.........x + 4

it has an asymatope at y=x.. now how do i know how to find that???

#### Lazarus

##### Retired
You use the normal procedure for finding asymptotes - take the limit as x approaches infinity. Let's see if I can remember how to do this. I'll use oo for infinity.

....lim..........x^2
x -> oo......x + 4

Divide by the highest power of x in the denominator.

-----------> (x^2)/x
................x/x + 4/x

----------->......x......
....................1 + 4/x

Substitute x for infinity, and remember that anything (except zero and infinity) over infinity is zero.

----------->....oo...
....................1 + 0

-----------> oo

Remembering that x is infinity.

-----------> x

As x -> oo, y -> x.

#### Enigma17

##### New Member
thanks lazarus... i THINK i'll be able to do it now...

#### Big Willy

##### Cabra Junkie RULES
Originally posted by Enigma17
Asamatopes
Nice spelling.
But anyeays, i have no trouble with it... except for working out what happens to f(x) when x--> negative infinity

#### Minai

##### Alumni
thats called oblique asymptotes

#### Big Willy

##### Cabra Junkie RULES
Originally posted by MinAi
thats called oblique asymptotes
u wernt talking about my reply were you?.. coz the obliques occures when x-->y

#### Weisy

##### the evenstar
teachers are usually good at explaining such things...

otherwise most textbooks (cambridge, fitz, MIF) have a 'curve sketching menu' which tells you exaclty how to sketch such graphs