horizontal asymptotes (1 Viewer)

[snow726]

Does this work for all functions? and limits?

Dividing the coefficient of x to the highest power (if the power is the same for both the numerator and denominator) gives the horizontal asymptote
When the power of x is greater in the denominator than in the numerator --> y=0

e.g. x^2+1/2x^2-1 = 1/2
therefore, y=1/2 is the horizontal asymptote

I want to confirm that this works for all functions and also limits (x --> ∞) just in case because i havent seen it in the cambridge textbook

thanks so much

 

[yanujw]

Does this work for all functions? and limits?

Dividing the coefficient of x to the highest power (if the power is the same for both the numerator and denominator) gives the horizontal asymptote
When the power of x is greater in the denominator than in the numerator --> y=0

e.g. x^2+1/2x^2-1 = 1/2
therefore, y=1/2 is the horizontal asymptote

I want to confirm that this works for all functions and also limits (x --> ∞) just in case because i havent seen it in the cambridge textbook

thanks so much

Yep all of that is correct. In terms of being asked to show working out, the method is to divide by all terms by the highest power and then show that terms of the form 1/x, 1/x^2 etc. all go to 0 as x goes to infinity, hence the only coefficients that remain are that were part of the highest power terms.

 

[smiley_riley]

Does this work for all functions? and limits?

Dividing the coefficient of x to the highest power (if the power is the same for both the numerator and denominator) gives the horizontal asymptote
When the power of x is greater in the denominator than in the numerator --> y=0

e.g. x^2+1/2x^2-1 = 1/2
therefore, y=1/2 is the horizontal asymptote

I want to confirm that this works for all functions and also limits (x --> ∞) just in case because i havent seen it in the cambridge textbook

thanks so much

I think it works generally, but not for all functions.
I assume you're talking about dividing the x^2/2x^2 giving 1/2.

I think it's because no matter what other values of x you have lower than the highest degree, at bigger numbers something like an x^2 will be worth significantly more than even like a 9999x would be.
And so taking the limits (And this is not solid math theory but my blah blah explanation) of f(x)/g(x), the only thing that will play any sort of effect will be the highest degree parts of f(x) and g(x) and so you can just divide them.

And of course there may be vertical asymptotes you have to worry about if you can find a zero for the denominator

Although I did some graphing on desmos and something like (x^3)/2x-2 doesn't fit a y = x^2 asymptote exactly
What does give a proper asymptote is by doing long division of the numerator divided by the denominator and the resulting quotient will be your asymptote graph, but you'd only really need to worry about that when the degree of the numerator is 2 or more than the denominator I think.

My real advice is that if you're doing HSC Maths there is a 99% chance you will never see a question with a sloped asymptote. So just dividing x^2/2x^2 is perfectly fine.

 

[SMARTYPANTS_SINGH]

if you're not too keen to know the math behind this, remember these rules and I'm pretty sure this works for every hyperbola / graph that’s meant to have a asymptote in it:

if Deg Numerator < Deg Denominator -----> HA = 0

if Deg Numerator = Deg Denominator -----> HA = leading coefficient of the numerator / leading coefficient of denominator

if Deg Numerator > Deg Denominator -----> HA is oblique and you have to do long division to find the basic equation of the HA (Note: don't worry about the remainder but as long as you have the main equation after dividing that should work as your HA)


if this doesn't make sense its essentially what you said and dw about oblique asymptotes they are very rare In questions unless you go Du in which case you'll be absolutely bombarded with them when you do sketching with derivatives.

 

[snow726]

thank you everyone