VCE Methods Question Thread!

[Stick]

Personally, I'd tackle that one using addition of ordinates (oh how I hate addition of ordinates though... XD).

[b^3]

You can do it by addition of ordinates, or you can split it up using a hybrid function, but you'll have three sections of the graph depending on whether whats inside each of the modulii is positive for both, negative for both or the opposite sign. Really, do it whichever way you feel more comfortable with.

Then it's easy to just plot those curves for the appropriate domain.

[Bluegirl]

EDIT: Stick has a different method above, either will work. Just choose the one you feel more comfortable with.

I understood your method better, but it was interesting to see it done differently by Stick

Thanks, both of you!

[Bluegirl]

Methods is going to be the death of me next year. Omg. It's already frustrating me because I keep getting stuck

If x^2= a(x+2)^2 + b(x+2) + c for all values of x, find the values of a, b and c.

[b^3]

Expand out the expression, then equate whats in front of each term, so equate coefficients  of the term, the coefficients of the terms, and the coefficients of the constant terms . Then you'll have three equations which you can then solve for , and .

It's basically the same method Stick used above, so to refer to that if you get stuck

[Stick]

Methods is going to be the death of me next year. Omg. It's already frustrating me because I keep getting stuck

If x^2= a(x+2)^2 + b(x+2) + c for all values of x, find the values of a, b and c.

Alright, I'll do the same sort of equating method as I did before.

Look if you get stuck

(expanding)

(expanding)



(since there is no x term on the left hand side, the x terms on the right hand side must cancel out)

(sub in a=1)



(since there is no constant on the left hand side, the constant terms on the right hand side must cancel out)

(sub in a=1 and b=-4)

[Bluegirl]

Alright, I'll do the same sort of equating method as I did before.

Look if you get stuck

(expanding)

(expanding)



(since there is no x term on the left hand side, the x terms on the right hand side must cancel out)

(sub in a=1)



(since there is no constant on the left hand side, the constant terms on the right hand side must cancel out)

(sub in a=1 and b=-4)

Before looking I managed to work out a=1 and c=4 (with some completely whacked method that was probably a fluke) but couldn't get b.

Thankyou so much.
I'm going to struggle so much next year. I was about to throw my books before it made me that frustrated. Haha. I swear I don't have anger issues.. Only when it comes to math..

[datfatcat]

Also, my suggestion is that after finding the values of a,b and c, write down these values again at the end. Examiners should read every line of your working out but sometimes they don't really care what you wrote as long as your answers are correct. Writing the values of a, b and c again  at the end prevents them looking for your answers (just something my methods teacher used to say)

[lzxnl]

Alright, I'll do the same sort of equating method as I did before.

Look if you get stuck

(expanding)

(expanding)



(since there is no x term on the left hand side, the x terms on the right hand side must cancel out)

(sub in a=1)



(since there is no constant on the left hand side, the constant terms on the right hand side must cancel out)

(sub in a=1 and b=-4)

For this particular problem, I'd sub in u=x+2, x=u-2, so we have (u-2)^2=au^2+bu+c
You can do this one quite easily (:

Another way is to let x=-2. This way, both brackets on the right hand side drop off and you get c=4.
Then, notice that the only coefficient of x^2 on the right hand side is a, so to match the x^2 on the left, a=1.
Finally, sub in some random value for x to find what b is, into x^2=(x+2)^2+b(x+2)=4

[Bluegirl]

Another way is to let x=-2. This way, both brackets on the right hand side drop off and you get c=4.
Then, notice that the only coefficient of x^2 on the right hand side is a, so to match the x^2 on the left, a=1.
Finally, sub in some random value for x to find what b is, into x^2=(x+2)^2+b(x+2)=4

That's what I did first and how I got a and c. But then I couldn't get b haha.

Thanks everyone!

[Nato]

I understand the whole intuition behind odd/even functions, but when presented with a question, how should i show which one it is.
In other words, what should my working out look like?

for example

[Snorlax]

I understand the whole intuition behind odd/even functions, but when presented with a question, how should i show which one it is.
In other words, what should my working out look like?

for example

first time using latex. Hope it works?

[Strawberrry]

Can anyone explain what this is used for?
logb (N) = loga (N) / loga (b)

Thanks 

[psyxwar]

Can anyone explain what this is used for?
logb (N) = loga (N) / loga (b)

Thanks 

change of base formula. I'm not sure what it's uses are now since the CAS does everything, but on the scientific it lets you covert between bases (as the scientific usually only has base e and base 10)

[Kuroyuki]

change of base formula. I'm not sure what it's uses are now since the CAS does everything, but on the scientific it lets you covert between bases (as the scientific usually only has base e and base 10)

It could a tech free question asking to simplify and express a log with a different base.
In that case the change of base formula would be used?