Asymptote (1 Viewer)

[frenzal_dude]

Hi,
f(x) = (x^2 - x + 1)/(x - 1)

I tried to sketch this, I worked out the limit as x --> infinity, and found that it is f(x) = x - 1, but when you sub in x = 0, you get f(0) = -1, but according to the asymptote this shouldn't be allowed right?

When I sketched the graph I saw that f(x) passes through the point (0,-1), would this point simply be discontinuous?

 

[Trebla]

The asymptote is actually y = x. Note that:
y = (x² - x + 1) / (x - 1) = [x(x - 1) + 1] / (x - 1) = x + 1 / (x - 1)

 

[frenzal_dude]

The asymptote is actually y = x. Note that:
y = (x² - x + 1) / (x - 1) = [x(x - 1) + 1] / (x - 1) = x + 1 / (x - 1)

Thanks for your help.

If you try to find the limit as x approaches infinity, isn't it true to divide everything by the highest power of x in the denominator? In which case y = x-1? I'm just wondering how that is wrong.

 

[hscishard]

Thanks for your help.

If you try to find the limit as x approaches infinity, isn't it true to divide everything by the highest power of x in the denominator? In which case y = x-1? I'm just wondering how that is wrong.

Well..in year 11 functions, we find the asymptote by dividing by the highest power in the fraction. But clearly this isn't a horizontal asymptote.

I've never seen a y=x asymptote in a maths ext 2 book. And I've read many.

Lmao excuse my ignorance
EDIT: Oh how cool! I just read about the oblique asymptote.

So first:

Divide x (highest power) on both num and denom

As x--> infinity, y--> x
Since 1/x = 0

The y=x asymptote happens when the power of x is greater than the power of x in the denom.

That's all the book has provided me.

But for the last part, I'm thinking that only when the power of x is greater than the denom by 1. If it was 2 it would be a parabola asymptote?????

Hope I somehow helped

 

[Trebla]

Thanks for your help.

If you try to find the limit as x approaches infinity, isn't it true to divide everything by the highest power of x in the denominator? In which case y = x-1? I'm just wondering how that is wrong.

If you were to divide numerator and denominator by the highest power of x first you get:
y = (1 - 1/x + 1/x²) / (1/x - 1/x²)
This approaches infinity as x goes to infinity (since the denominator approaches zero)

The reason it failed is because the numerator has a higher degree than the denominator. If you want to investigate HOW the curve approaches in infinity with a bit more detail then just apply polynomial long division to simplify that expression (or other manipulative means) such that the degree of the numerator is always strictly less than the degree of the denominator.

 

[Omnipotence]

Well..in year 11 functions, we find the asymptote by dividing by the highest power in the fraction. But clearly this isn't a horizontal asymptote.

I've never seen a y=x asymptote in a maths ext 2 book. And I've read many.

Lmao excuse my ignorance
EDIT: Oh how cool! I just read about the oblique asymptote.

So first:

Divide x (highest power) on both num and denom

As x--> infinity, y--> x
Since 1/x = 0

The y=x asymptote happens when the power of x is greater than the power of x in the denom.

That's all the book has provided me.

But for the last part, I'm thinking that only when the power of x is greater than the denom by 1. If it was 2 it would be a parabola asymptote?????

Hope I somehow helped

Yep, it is an oblique asymptote. All you do is long division.

 

[fullonoob]

quoting from hscishard
"Divide x (highest power) on both num and denom"
on your maths equation 1 / (x-1) is alrdy in fractional form, so if you limit that to infinite it would approach zero. What is the need of dividing by x again. Isn't that just complicating things :nod

 

[frenzal_dude]

quoting from hscishard
"Divide x (highest power) on both num and denom"
on your maths equation 1 / (x-1) is alrdy in fractional form, so if you limit that to infinite it would approach zero. What is the need of dividing by x again. Isn't that just complicating things :nod

Do you mean, we can look at x + 1/(x-1) and see that as x approaches infinity, 1/(x-1) will go to 0? That's what I thought aswell.

 

[fullonoob]

yeah that cos why put a fraction in another fraction, no point

 

[nikkifc]

Hi,
f(x) = (x^2 - x + 1)/(x - 1)

I tried to sketch this, I worked out the limit as x --> infinity, and found that it is f(x) = x - 1, but when you sub in x = 0, you get f(0) = -1, but according to the asymptote this shouldn't be allowed right?

When I sketched the graph I saw that f(x) passes through the point (0,-1), would this point simply be discontinuous?

Um, did I read correctly that you tutor Ext 1 mathematics? oO