Are my asymptotes wrong? (Ext. Maths) (1 Viewer)

[Guernica]

The question is to sketch y = x + 1/x
This is what I've done so far... are my asymptotes wrong? I think they are..

 

[Slidey]

y = x + 1/x

Vertical asymptote at x=0
Oblique asymptote at y=x

Can you guess when y=x^2+1/x will display "asymptotic" behaviour?

Here is my graph of y=x+1/x

 

[sabdow]

Slide Rule's one is correct.

If you deduce it's behaviour (Take large and larger values for x) while doing the limit, you'll find out that as x gets larger it starts to behave like y = x (a straight line passing through the centre).

That's why it looks like that incase anybody is wondering.

 

[Guernica]

y = x + 1/x

Vertical asymptote at x=0
Oblique asymptote at y=x

Can you guess when y=x^2+1/x will display "asymptotic" behaviour?

Here is my graph of y=x+1/x

*cough* I see..

 

[KFunk]

Generally if you're working with a graph involving 1/f(x) then where f(x)=0 the graph of 1/f(x) will have an asymptote.

Another thing to consider in terms of oblique asymptotes is the behaiviour of f(x) as x approaches infinity

As x ---> infinity , 1/x ---> 0 and x + 1/x ---> x
Hence,
as x ---> infinity , y ---> x

 

[Slidey]

Use this for comparison. Basically it's called addition of ordinates and you just superimpose the two graphs (y=x and y=1/x). You will cover this thoroughly in 4unit.

 

[Guernica]

We haven't done anything about oblique asymptotes... I wonder why the teacher set this question for homework.

 

[Slidey]

Possibly because you're about to do work no oblique asymptotes?

It's not hard. Do as KFunk said:

Determine what y approaches as x approaches infinity.

E.g.:

for y=x²+1/x
as x approaches infinity, 1/x approaches 0, so y must approach x²

So basically the graph of y=x²+1/x will look like y=x² except around the origin where it will act like 1/x

 

[Guernica]

Mm, yeah, we've done all the working of what happens if x approaches infinity etc... I didn't know that it could do stuff like this though.
Thanks for helping, you guys!

 

[Trebla]

y = x + 1/x

A quick way to find the asympote is by this method: (Don't know if you've learnt it or your school approves of it though...)

Make it in the form (x - a)(y - b) = c
Then, you have x = a and y = b as your asymptotes. The c will tell you about the steepness and position of the hyperbola.
So:

y = x + 1/x
=> y - x = 1/x
=> x(y - x) = 1
or: (x - 0)(y - x) = 1
.: asymptotes are x = 0 and y = x

 

[YBK]

Yep, that's right... i tried the problem and got the same result as slide rule..

to get the vertical asymptote you replace by 1/x

so it will be like this

y= (x^2(1/x) + 1/x) / x(1/x)

Therefore y = 1/0
y = 0

And since the numerator is one power higher than the dinominator then it has an oblique asymptote.

Also x not equal to 0


If you don't know how to graph it, just replace a few values x and you'll see the trend for y.

for eg. if x = 5
y = 5^2+1/5
= 26/5

 

[shafqat]

y = x + 1/x

A quick way to find the asympote is by this method: (Don't know if you've learnt it or your school approves of it though...)

Make it in the form (x - a)(y - b) = c
Then, you have x = a and y = b as your asymptotes. The c will tell you about the steepness and position of the hyperbola.
So:

y = x + 1/x
=> y - x = 1/x
=> x(y - x) = 1
or: (x - 0)(y - x) = 1
.: asymptotes are x = 0 and y = x

Just to add a note of explanation to that method:

What we're arguing here is that as the product is 1, if one factor approaches +/-infinity, the other factor approaches 1/+-infinity = 0.

So consider x. As it approaches +-infinity, y-x, the other factor, must approach 0. That is, y-->x. So a point(x,y) gets as close as we like to the line y = x. That is y = x is an asymtote.

Similarly near x = 0, y - x = 1/x approaches +-infinity. And so x = 0 is a vertical asymptote.

This is the general method used to find the asymptotes of hyperbolas in ext2 conics.

 

[YBK]

Ummm... to clarify the method we were taught, u just divide by the highest power of x to get the vertical asymptote.

For example:
x/x^2

The vertical asymptote is 0/1 = 0
And obviously the horizontal asymptote is 0, because you can not divide by zero.

I think it's much easier/faster than trebla's method.

Another example:
x^2 - x^3/ x^3 - 1

The vertical asymptote is -1
Horizontal asymptote is, obviously 1 because x^3 - 1 can be factored into (x - 1) (x^2 + x + 1)

actually, u dont even need to factor the dinominator to get the horizontal asymptote, it's 1 cause 1 ^ 3 - 1 = 0 and denominator can't be 0.

ah crap, we have a maths exam on wednesday...

 

[shafqat]

That's the best method for vertical/horizontal asymptotes. But what about for oblique asymptotes? Trebla's method works for those.

 

[YBK]

That's the best method for vertical/horizontal asymptotes. But what about for oblique asymptotes? Trebla's method works for those.

if the top power is 1 higher than the bottom power then u can deduce that it has an oblique asymptote

for example:

y = x^2 + 1/x


the top power is one higher than the bottom, therefore it has a oblique asymptote.

Well, at least i think that works... correct me if i'm wrong.

 

[shafqat]

try x^2 - y^2 = 9, or something like that. i guess its a little advanced for prelim, but if you do 4u you'll need to use a method like trebla's later on.

 

[YBK]

try x^2 - y^2 = 9, or something like that. i guess its a little advanced for prelim, but if you do 4u you'll need to use a method like trebla's later on.

is it a circle?

 

[shafqat]

for a circle you'd need a + instead of the -
its a rectangular hyperbola.