VCE Methods Question Thread!

[Shadowxo]

Ok.. it's not allowing me to attach picture so I need to write out the question .-.

Anyways this is what the question asks

After the rebellion was successful in overthrowing the  empire, the number of Jedi  that were discovered  and trained increased according to the rule  J=alog2(t)+b, where a and b are constants and t is the time in years after he battle of endor.
Two years after the battle of Endor,  The new Republic was formed, and Luke and leila were the only two known jedi.
Thirty years later, after Luke established a jedi school on yavin four, there were 82 known jedi in existence

A) determine the values of a and b
B) sketch the graph against J against t for the first 50 years of the new republics existance.

Ok so I need help with A)
So I'm confused. This is what I did. Made simultaneous equations of
2=alog2(2)+b
And
82=alog2(32)+b

Then I got a=20 and b =-18 then subbed in actual rule. But when I sketched this it was giving me wrong numbers when x=2.11 it is giving y=3.55. (That is wrong) y is supposed to be equal to 2. What am I doing wrong. Thanks

In your question you wrote 30 years then subbed in 32 - I'm assuming this is a typo and you meant 32 years?
So 2=alog₂2+b = a+b
82=alog₂32+b = 5a+b
82-2=5a-a
4a=80 a=20 as you found out
a+b=2 so b= -18

J is only equal to 2 when t=2, not when t=2.11 (not sure where that came from?)
Your answers for a) should be right provided it's 32 years and not 30

[FedererFan]

Consider the function f(x)=(x-1)^2(x+2)
State the transformations required to map the graph of y=f(x) into the graph of y=h(x) where h(x)=Af(x-h)

Hey, i am not very good with transformation. Im not sure how to do this question. Thanks

[zhen]

Consider the function f(x)=(x-1)^2(x+2)
State the transformations required to map the graph of y=f(x) into the graph of y=h(x) where h(x)=Af(x-h)

Hey, i am not very good with transformation. Im not sure how to do this question. Thanks

Someone should correct me if I'm wrong because I'm not that good at transformations and always stuff up the order, but I'm pretty sure that you dilate by a factor A away from the x-axis and then translate h units in the positive x-direction. If you're stuck on a question about translations, just think about what would happen if the function was If you change it to you're dilating by a factor A from the x-axis. Also if you're changing it to you're changing the turning point to (h,0), so you're moving it h units in the positive x direction. Hope that helps. 

[QueenSmarty]

Hello, may I please get some help on these questions?
1. State the transformations that map f(x)=(x-1)^2(x+2) onto y=f(x/n)+k
2. Using the product rule, find the derivative of f(x)=(x-a)^n(x+b), hence find the stationary points

[zhen]

Hello, may I please get some help on these questions?
1. State the transformations that map f(x)=(x-1)^2(x+2) onto y=f(x/n)+k
2. Using the product rule, find the derivative of f(x)=(x-a)^n(x+b), hence find the stationary points

1. Dilate by a factor n from the y-axis and translate k units in the positive y direction. To explain this I'll steal Sine's explanation. Hopefully they don't mind.

So when you are dilating by factor 2 from the y-axis, what are you doing? You are replacing each x with x/2.

When you are dilating by factor 2 from the x-axis, you may say you are sticking a 2 in front of the equation but that is wrong. You are actually replacing y with y/2 which yields y/2 = x^2 this gets you to y = 2x^2

2. f(x)=(x-a)ⁿ(x+b)
f'(x)=(x-a)ⁿ+n(x-a)^n-1(x+b)
When f'(x)=0, x=a
f(a)=0, so (a,0) is a stationary point. This is only if n doesn't equal 0 or 1.

There are probably more stationary points, but I can't find them right now. I'll get back to it later, but someone should probably finish off 2.

Edit: f'(x)=(x-a)^n-1(n(x+b)+(x-a))
f'(x)=(x-a)^n-1(nx+nb+x-a)
So nx+nb+x-a=0
nx+x=a-nb
x(n+1)=(a-nb)
x=(n+1)/(a-nb)
Here's what I got for the second turning point. I'm not sure about it, so someone should correct me if I'm wrong.

[QueenSmarty]

Thank you zhen!
One more question, describe the effect an odd or even value of 'n' has on the function f(x)=(x-1)^n(x+2).
This is what I have so far, although I'm not so sure that it really answers the question:
-as odd values of 'n' increase, the graph becomes narrower as the graph is dilated from the y-axis by a higher factor
-even values of 'n' have a stationary point above the x-axis, whereas odd values of 'n' have a stationary point below the x-axis

[zhen]

Thank you zhen!
One more question, describe the effect an odd or even value of 'n' has on the function f(x)=(x-1)^n(x+2).
This is what I have so far, although I'm not so sure that it really answers the question:
-as odd values of 'n' increase, the graph becomes narrower as the graph is dilated from the y-axis by a higher factor I don't think this answers the question
-even values of 'n' have a stationary point above the x-axis, whereas odd values of 'n' have a stationary point below the x-axis

I think one more thing to add is that when n is odd there is a stationary point of inflection at x=1. When it's even it touches the x axis at x=1 and goes back up. Someone else should check answer this since I don't know what the question is asking.

[FedererFan]

If there is a turning point on g(x) at (2,3) find all possible values of n and k
g(x)=(x-1)^2(x+2)+k
       ------     -----
         n            n

[geminii]

Say if there was a graph y=m(x-c)^2(x-d), could m be negative? The question states to sketch this graph and that c does not equal d, and that m<0. So how would this negative dilation work?? 

[deStudent]

Say if there was a graph y=m(x-c)^2(x-d), could m be negative? The question states to sketch this graph and that c does not equal d, and that m<0. So how would this negative dilation work?? 

It just means it'll also be reflected in the x-axis.

[Syndicate]

If there is a turning point on g(x) at (2,3) find all possible values of n and k
g(x)=(x-1)^2(x+2)+k
       ------     -----
         n            n

Hint:
eqn1. \(\frac{dy}{dy}\) = 0 (differentiate the function and substitute x = 2)
eqn2. In the original function substitute x = 2, y = 3.

Now that you have a consistent system of equations, you can solve for n and k simultaneously

[Cyka]

Does anyone here have any methods application SACS that they are willing to share (Based on Ch1-10 of Cambridge textbook - functions, graphs, applications of differentiation)?

[zhen]

Does anyone here have any methods application SACS that they are willing to share (Based on Ch1-10 of Cambridge textbook - functions, graphs, applications of differentiation)?

I don't have any methods application SACs to share, since my next methods SAC is on related rates and a bit of implicit differentiation, which is basically specialist, so it's definitely not going to be related to your methods SAC. But, I would recommend you to do the textbook chapter reviews if you haven't already done them and if you have the other textbooks, then do questions from them too for extra practice. Also, if you can get a hold of some old practice exams (not VCAA past exams, since you want to save them for exam time), then I recommend you to pick out specific questions related to what the topics you've covered. For my specialist SAC I did questions from old practice exams from like 2006 and 2007 that were relevant to my SAC. Chances are during exam revision time, you're not going to look at these old exams anyway, since they're old and out of date, but it can still help for SAC revision. Hope that helps and good luck for your SAC.

[TheCommando]

Hi, i keep getting question 4 ii wrong
For eg. 4a ii
4cii how do i do theese

[Shadowxo]

Hi, i keep getting question 4 ii wrong
For eg. 4a ii
4cii how do i do theese

Well, for questions like these, f o g is defined where the range of g is a subset of the domain of f (aka f(x) has a valid input). You then substitute in these  values into f(x) to find the range of f(g(x)), and find the values of x that give these values to find the maximal domain.
eg 4a ii
f(x) = 3sin(2x) ,  g(x) = x²
for f o g:
range of g(x) = [0,infinity) ,  domain of f(x) = (-infinity, infinity), so f o g exists where range of g(x) is [0,infinity), so domain of f(x) becomes [0,infinity) also. This results in a range of [-3,3] (plugging in 0 to infinity into f(x) ). You get [0,infinity) from g(x) when x is (-infinity,infinity) so that is the maximal domain.
For g o f you follow the same process.

Hope that helps, let me know if you want me to do another example / one's particularly difficult