VCE Methods Question Thread!

[TrueTears]

ah okay, i see what you mean, tbh i don't have too much validity in the first method, however the second method is quite valid, what i mean is, we have to manipulate f(x) into satisfying f(-x)=f(x), now the hyperbola in the question, p+(q-pr)/(x+r), is clearly not even. However only one part of any hyperbola is even, that is it's horizontal asymptote, so we pretty much "delete" off every part of the hyperbola and just leave it's horizontal asymptote intact, in other words p+(q-pr)/(x+r) just becomes p since we delete the (q-pr)/(x+r) component, so f(x) = p.

[yawho]

ah okay, i see what you mean, tbh i don't have too much validity in the first method, however the second method is quite valid, what i mean is, we have to manipulate f(x) into satisfying f(-x)=f(x), now the hyperbola in the question, p+(q-pr)/(x+r), is clearly not even. However only one part of any hyperbola is even, that is it's horizontal asymptote, so we pretty much "delete" off every part of the hyperbola and just leave it's horizontal asymptote intact, in other words p+(q-pr)/(x+r) just becomes p since we delete the (q-pr)/(x+r) component, so f(x) = p.

So you are saying the asymptote is part of the hyperbola.

[TrueTears]

well it's in the equation why wouldn't it be? just happens p = horizontal asymptote

[yawho]

What is the meaning of an asymptote?

[Phy124]

A line that continually approaches a given curve but does not meet it at any finite distance.

For example, a graph of y = 1/x has asymptotes at x=0 and y=0 because the graph approaches, yet never touches these "lines" (of x=0, y=0) or vice versa

Asymptotes are usually represented by dotted lines.

[TrueTears]

What is the meaning of an asymptote?

what Phy124 said, however even if you aren't convinced that asymptotes are 'part' (although I'm not sure of your definition of "part") of the hyperbola there is another way to think about it [btw you can argue both ways that asymptotes are and aren't part of the graph, personally i am not sure of what the mathematical convention is, however lim_x-> infinity p+(q-pr)/(x+r) = p, so you can assume it does "touch" p at infinity; but then the question arises as to how you define infinity]

but anyways, consider p+(q-pr)/(x+r) as an equation rather than a graph, clearly the geometric representation of an even equation is not satisfied by the aforementioned asymmetric equation, the only way we can make the equation geometrically symmetric is to cut out the (q-pr)/(x+r) component, so in either way, we are let with f(x) = p

[yawho]

A horizontal asymptote of a function is a straight line that the function approaches as x approaches +/- infinity, and it is not part of the function. So your explanation fell apart when you based it on a false assertion.

"consider p+(q-pr)/(x+r) as an equation rather than a graph" I thought an equation has an equal sign in it.
Exactly what did you mean by 'cut out the (q-pr)/(x+r) component'?

[TrueTears]

how do you define something to be "part" of a function? Who says an asymptote isn't "part" of a function?

wiki seems to consider asymptotes to be 'part of' the function, http://en.wikipedia.org/wiki/Asymptote

In some contexts, such as algebraic geometry, an asymptote is defined as a line which is tangent to a curve at infinity.

like i said, the argument breaks down into your definition of infinity, which is pointless to digress into for a VCE question.

I mean cut out as in the only way to get an even function out of p+(q-pr)/(x+r) is to remove (q-pr)/(x+r). If you want to be formal, I guess you can say I 'set' (q-pr)/(x+r)=0

"consider p+(q-pr)/(x+r) as an equation rather than a graph" I thought an equation has an equal sign in it.

obviously f(x) = p+(q-pr)/(x+r)

[yawho]

the link does not consider asymptotes to be 'part of' the function

[TrueTears]

define your mathematical definition of "part of"; please enlighten me.

and yes it does.

[kamil9876]

You guys are making a meal of a simple question. Here is a more tangible explanation of TT's solution (without using controversial words like 'infinity', 'part of', 'cut' etc.):

The only hyperbola that's an even function is constant, so is constant. Now but clearly so

[yawho]

You guys are making a meal of a simple question. Here is a more tangible explanation of TT's solution (without using controversial words like 'infinity', 'part of', 'cut' etc.):

The only hyperbola that's an even function is constant, so is constant. Now but clearly so

'The only hyperbola that's an even function is constant' Don't you think it sounds strange.
In fact tt finally arrived at the right explanation.
tt should be credited in writing f(x) = p+(q-pr)/(x+r), i.e. the sum of an even function p and a function (q-pr)/(x+r) which is neither even nor odd. For f(x) to be even, (q-pr)/(x+r) must be zero, i.e. q-pr=0 and f(x)=p.

[TrueTears]

Yes, I never intended to butcher the question, I assume my overview explanation would suffice. A VCE methods way is just to formalise what I said before, f(x) = p+(q-pr)/(x+r), let p(x) = p and h(x) = (q-pr)/(x+r), so f(x) = p(x)+h(x), since the sum of 2 even function has to be even, then for f(x) to be even, p(x) has to be even and h(x) has to be even. p(x) is already even, however h(x) is not, to make h(x) even we set q = pr, such that h(x) = 0. Hence f(x) = p.

EDIT: yes beaten, that's what I meant

[yawho]

You guys are making a meal of a simple question. Here is a more tangible explanation of TT's solution (without using controversial words like 'infinity', 'part of', 'cut' etc.):

The only hyperbola that's an even function is constant, so is constant. Now but clearly so

I thought you said not using infinity.

[kamil9876]

'The only hyperbola that's an even function is constant' Don't you think it sounds strange.

It's true, I'm assuming that's what he was referring to when he was talking about "hyperbola's are not even in nature" and then talking about the 'geometry of a hyperbola'.

In fact tt finally arrived at the right explanation.

So what is your problem if you are satisfied with his explanation?

I thought you said not using infinity.

Sure, I didn't use any statement like "at infinity...". Limits can be defined in a precise way without using the word infinity at all.