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Author Topic: VCE Methods Question Thread!  (Read 4803480 times)  Share 

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sogreatsosad

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Re: VCE Methods Question Thread!
« Reply #19080 on: February 15, 2021, 09:42:05 pm »
0
https://imgur.com/a/CCJ1vbl

Can someone solve the above? Teacher gave it to us as a "challenge". I tried abc-(a-2)(b-2)(c-2)=(a-2)(b-2)(c-2) to solve for b and c (having already subbed a for 8 )

I get b=6(c-2)/(c-6) and c=6(b-2)/(b-6)

Then I do 8bc but I sub in b and then derive to find the maximal, but I don't find an integer for b or the volume


fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19081 on: February 15, 2021, 10:07:11 pm »
+5
Hint: You don't need to explicitly find values for b and c straight away, though it might seem counterintuitive at first. Answer in the spoiler, but give it another go first :)

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Corey King

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Re: VCE Methods Question Thread!
« Reply #19082 on: February 16, 2021, 06:10:06 pm »
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Hey guys,
Having trouble with working through this one.

Q 14 :)

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19083 on: February 16, 2021, 06:23:52 pm »
+5
Hint: For any parabola \((x-a)(x-b)\) where \(a, \ b\) are the roots of the parabola, the largest domain for which the parabola has an inverse is one of \([\frac{a+b}{2}, \infty)\) or \((-\infty, \frac{a+b}{2}]\). What does this tell you about the value of b?

Hint: The inverse function for one of the 'branches' of the parabola is some square root function. Is there a correlation between the branch of the parabola chosen and the sign of the square root function? Try rewriting the function as \(y=x^2+4x\) and noting that the inverse function is a subset of the equation \(x=y^2+4y\).

Hope this helps :)
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19084 on: February 16, 2021, 08:45:47 pm »
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I feel like Ive been spending all afternoon/evening trying to understand something simple.

I don't see how you got (a+b)/2. Or why that's true

Also, I can see how a square root graph has a 'root'. If a parabola has a root, would it not be one root with two 'sprouts', not two (a and b). Im not sure why you pointed this out specifically

No idea what that tells me about b.

Also, all I know how to do is to substitute y for x and do the algebra bit. But I didnt know how to isolate y and so got this far:

f^-1(x)=y(y + 4)


arnavg2207

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Re: VCE Methods Question Thread!
« Reply #19085 on: February 16, 2021, 08:57:09 pm »
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Can someone please help me with this!

There are approximately five times as many magpies as currawongs in a certain area.
If the population of currawongs increases at a rate of 12% per annum while that of
the magpies decreases at 6% per annum, find how many years must elapse before the
proportions are reversed, assuming the same rates continue to apply.

(Q9 EX 5I)

Thanks!
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2022: Legal 3/4 Physics 3/4 Chemistry 3/4 English 3/4

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19086 on: February 16, 2021, 10:31:18 pm »
+6
Can someone please help me with this!

There are approximately five times as many magpies as currawongs in a certain area.
If the population of currawongs increases at a rate of 12% per annum while that of
the magpies decreases at 6% per annum, find how many years must elapse before the
proportions are reversed, assuming the same rates continue to apply.

(Q9 EX 5I)

Thanks!

Welcome to the forums!

Let the starting number of magpies be \(5x\), and the starting number of currawongs be \(x\). Then, after \(n\) years, the number of magpies will be \(5x(1-0.06)^n\) and the number of currawongs will be \(x(1+0.12)^n\). We want to find when the number of currawongs will be 5 times that of the magpies or more - ie what is the smallest value of \(n\) that satisfies the inequality \(\frac{x(1+0.12)^n}{5x(1-0.06)^n} > 5\) ie. solve this inequality for \(n\) - \(\left(\frac{1.12}{0.94}\right)^n > 25\). Try this out :D

I feel like Ive been spending all afternoon/evening trying to understand something simple.

You haven't, I probably started with too many assumptions :(

I don't see how you got (a+b)/2. Or why that's true

The largest domain for which the parabola has an inverse lies between positive/negative infinity and the vertex. The vertex is the maximum if the parabola is a 'cap' or a minimum if the parabola is a 'cup'. More importantly for our purposes, the parabola is symmetrical across the vertical line that passes through the vertex. Since this is the case, its x-coordinate must be halfway between each of the roots a and b, hence (a+b)/2. The horizontal line test (if a function has some real y value, any real y value for which a horizontal line through that y value cuts through the function twice, it does not have an inverse across its entire domain - you have to restrict the domain for it to have an inverse) fails because of this symmetry. This symmetry automatically gives you the second point that fails the horizontal line test - in fact the parabola fails this test everywhere in the range of the function (except for the vertex).

Also, I can see how a square root graph has a 'root'. If a parabola has a root, would it not be one root with two 'sprouts', not two (a and b). Im not sure why you pointed this out specifically

I didn't use root in inverted commas because roots weren't intentionally metaphorical like 'branches' (which is a commonly used term, I think? but not formal). Roots are just another way of saying the solutions of the parabola, or if it suits you better the x-intercepts of the parabola. Hopefully things make more sense with this clarification - all parabolas have two roots (either two distinct roots, two duplicate roots or two complex roots (but this is a whole different story)).

No idea what that tells me about b.

You can put \(x^2+4x\) into the form \((x-a)(x-b)\) by factorising it. This will tell you that the roots of the equation are 0 and 4.
Since b is the smallest value for which \(g\) has an inverse, it follows that b has the same x value as the vertex ie. (a+b)/2 = (0+4)/2 = 2.

Also, all I know how to do is to substitute y for x and do the algebra bit. But I didnt know how to isolate y and so got this far:
f^-1(x)=y(y + 4)

The easiest method is probably completing the square.


Depending on your domain restriction (assuming it is restricted such that the domain is the largest for which g has an inverse) you now have to choose a sign for the root because choosing both means that this is not a function. For 'cup' parabolae if you choose the one that has an open interval in the positive x direction you choose the positive root, otherwise you choose the negative root. For 'cap' parabolae the opposite is true.

Hope this makes more sense
« Last Edit: February 16, 2021, 10:58:16 pm by fun_jirachi »
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19087 on: February 17, 2021, 12:16:08 am »
+4
My bad, I was thinking root may have meant something completely different.

You had me with roots for a bit, until here :P "Hopefully things make more sense with this clarification - all parabolas have two roots (either two distinct roots, two duplicate roots or two complex roots (but this is a whole different story))."

I know the x-intercept terminology :)

Finding the vertex makes sense. I got the x-intercepts but didnt know what to do with them, if anything :P

I forgot that inverse functions have to be one-to-one, and cannot be many-to-one. Since the inverse would turn the many-to-one into one-to-many.

We still haven't really learned why it can't be many to one. It doesn't seem too bad just having two inputs per output :P

Thank you for the effortful and detail response Jirachi. :)


Bluebird

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Re: VCE Methods Question Thread!
« Reply #19088 on: February 18, 2021, 10:45:50 pm »
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Hey AN!
Can I have some help with this question? I've tried putting into my CAS:
f(0.75)=3.1
f(1.5)=0
f(0)=0
but I don't know if my values are even right because my calculator just says false.
Thank you in advance!

1729

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Re: VCE Methods Question Thread!
« Reply #19089 on: February 19, 2021, 07:40:39 am »
+2
Hey AN!
Can I have some help with this question? I've tried putting into my CAS:
f(0.75)=3.1
f(1.5)=0
f(0)=0
but I don't know if my values are even right because my calculator just says false.
Thank you in advance!
I don't understand what you're doing to solve this problem, but this is how I would solve the problem.

These are the points you have: (0,0) (1.5,0) and (0.75,0.6)

Note: You know the value of c is 0 since the parabola passes the origin.

y = a(x^2 - 1.5x)
0.6 = a(0.75^2 - 1.5(0.75))
0.6 = a(0.5625 - 1.125)
0.6 = -0.5625a
a = -16/15
now you can do...
-16/15(x^2 - 1.5x) and distribute and that is your funtion

Also I think your error was where you thought (0.75,3.1) was a cordinate. Cordinate of vertex is actually (0.75, 0.6) because 3.1 - 2.5

Hope this helps
« Last Edit: February 19, 2021, 07:47:52 am by 1729 »

Corey King

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Re: VCE Methods Question Thread!
« Reply #19090 on: February 19, 2021, 09:29:30 am »
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Hey guys :)
Do any of you know if this old TI calculator will be able to cover all the functions needed for the current study design?

The Cat In The Hat

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Re: VCE Methods Question Thread!
« Reply #19091 on: February 19, 2021, 09:36:48 am »
+1
Hey guys :)
Do any of you know if this old TI calculator will be able to cover all the functions needed for the current study design?
I believe not. That's what I used to have before I inherited my brother's but he'd had to get the newer one for Methods in any case. I'd at least strongly advise against it since you'd not have any guide for some things and I don't think you can even do all the functions required. I'd advise you to replace it with a newer one instead.
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Re: VCE Methods Question Thread!
« Reply #19092 on: February 19, 2021, 10:00:27 am »
0
Damn, thought that might be the case :/
How new does it need to be? :P

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Re: VCE Methods Question Thread!
« Reply #19093 on: February 19, 2021, 11:08:19 am »
+1
Damn, thought that might be the case :/
How new does it need to be? :P
I'm afraid I don't know. Probably ask your teacher. And also look secondhand if you want not quite such a horrid price.
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Bluebird

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Re: VCE Methods Question Thread!
« Reply #19094 on: February 19, 2021, 03:19:21 pm »
+1
I don't understand what you're doing to solve this problem, but this is how I would solve the problem.

These are the points you have: (0,0) (1.5,0) and (0.75,0.6)

Note: You know the value of c is 0 since the parabola passes the origin.

y = a(x^2 - 1.5x)
0.6 = a(0.75^2 - 1.5(0.75))
0.6 = a(0.5625 - 1.125)
0.6 = -0.5625a
a = -16/15
now you can do...
-16/15(x^2 - 1.5x) and distribute and that is your funtion

Also I think your error was where you thought (0.75,3.1) was a cordinate. Cordinate of vertex is actually (0.75, 0.6) because 3.1 - 2.5

Hope this helps

Thank you! I get it now :)