A question about hyperbolic functions. (1 Viewer)

[V I Z E]

Hi guys,

So when it comes to finding horizontal/vertical asymptotes for hyperbolas, we have to show full working out despite the fact there are shortcuts with limits etc.

The thing is, we haven't done limits yet in 2U (at my school) so we're not expected to use it in the exam. Instead, we have to "split" the hyperbolic function.

E.g. Given the function f(x) = x/x-1, find asymptotes.

To split, we basically split the "x" up so it becomes
f(x) = (x-1+1)/(x-1), and then,
f(x) = (x-1)/(x-1) + 1/(x-1)
f(x) = 1 + 1/(x-1)

So then we look at the constant "1" and say that's the horizontal asymptote. Let x-1 = 0, then x = 1 is vert. asympt.


But if we're given a complicated hyperbola with both linear expressions on numerator and denominator:

e.g. f(x) = (2x+1)/(5-2x) or f(x) = (-7x+1)(x-6) or f(x) = (-1 + 5x)/(5-7x)

then how are we supposed to "split" it?

I'm really confused atm. Usually I just "look" at the leading co-efficients then I get the HA straight away. Likewise, all I do is let denominator = 0 to find VA.

Have an upcoming 2U exam. Some please come to my aid!

Thanks.

 

[braintic]

(i) (2x+1) / (5-2x) = [ -(5-2x) + 6 ] / (5-2x) = -1 + 6 / (5-2x)

(ii) (-7x+6) / (x-6) = [ -7(x-6) -36 ] / (x-6) = -7 - 36 / (x-6)

(iii) (-1 + 5x) / (5-7x) = [ (-5/7) (5-7x) +18/7 ] = -5/7 + 18 / [7(5-7x)]

Basically, throw the denominator up top, multiply it all by a number with the aim of getting the x term correct, then add/subtract what you need to get the constant term correct.


And a harder one: y=(x^2 - 2x + 3) / (x+3)

(x^2 - 2x + 3) / (x+3)
= [ x(x+3) - 5x + 3 ] / (x+3)
= [ x(x+3) - 5(x+3) +18 ] / (x+3)
= x - 5 + 18/(x+3)

So it has an oblique asymptote of y = x - 5

ALTERNATIVELY, if you know how to do polynomial division then do that. It is just a different way of setting out exactly the same process.

 

[Carrotsticks]

That's the general way to do it too, 'look' at the leading coefficients (this only works when the highest power of the denominator and numerator are the same).

But say you wanted to do something more complex, I'll give you an example.



Let us know if you don't understand how I did any of the above steps!

 

[V I Z E]

Got it. Thank you so much!!

It's just the manipulation. I did a few examples and the ones above mentioned so now I got the hang of it

 

[V I Z E]

Another question - absolute functions.

- if we square an absolute, say |x|^2, would |x|^2 = x? or |x| = x^2?

Just need some clarification.

 

[braintic]

Another question - absolute functions.

- if we square an absolute, say |x|^2, would |x|^2 = x? or |x| = x^2?

Just need some clarification.

You should be able to discover the answer yourself just by subbing in values for x.