VCE Methods Questions Thread

Billzene

snowflake Your answer wouldn't be considered correct, if the coefficient of x2 were 5 ie y = 1 / (5x2) then it'd be correct since that would be a dilation from the y-axis

snowflake

Billzene

Ohh ok but how do we whether to work out it’s dilation from the X or Y axis if it doesn’t say in the question?

Billzene

snowflake by recognition of the form y = a/(1/b(x-h))ⁿ + k

a = dilation factor from x axis
b = dilation factor from y axis
h = translation in +ve direction of x axis
n = power
k = translation in the positive direction of y axis

This is similar to turning point form of parabolas

snowflake

Billzene

thank youuu

chemistry1111

find values of a and b such that a(x+2)+b(x+3)=18x+8 for all values of x
if someone could help that would be great

beep_boop_booop

chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28

chemistry1111

thank you so much that makes sense

also this question
For the polynomial P(x)=(a+1)x³ + (b-7)x² +c+5, find values for a,b and c if P(x) has a degree 2, a leading coefficient of 3 and the constant term is -1

chemistry1111

when factorizing polynomials in methods unit 3/4, what method do most people use vcaa prefer, long division or equating coefficients

PizzaMaster

Do we need to buy the new Methods 3/4 book for the new study design (2023)?
I feel like the textbook questions are the same...

PizzaMaster

Can anyone plz help with this question (part b, c and d in particular):

https://imgur.com/a/7hzDLFw

jinx_58

PizzaMaster
Howdy, I did QCE Methods, so if I mess this up, I’m really sorry.

https://ibb.co/m0Q1xdV

For b) may I know what the answers are? I may be able to work backwards 🙂

I didn’t do d) because I wasn’t that confident in doing the question.

Hope this helps,
-jinx_58

PizzaMaster

Hi jinx_58,

Thank you so much for solving!
The answer to part b is Domain = [0, 6]; Range = [0, 9/2]
The answer to part c is 9/2

chemistry1111

The graph of y = 5/x -6 is reflected in the x-axis and
then in the y-axis. The equation of the final image is:

could someone help with this question

crumblycupcakes

chemistry1111

reflected in the x-axis
reflected in the y-axis

**I'm not quite sure if y = (5/x) -6 or if it's y = 5/(x-6) so I'll show the working for both

let f[x] = (5/x) -6

x reflection: -f[x] = f1[x] = -[ (5/x) - 6 ] = -( 5/x ) + 6
y reflection: f1[-x] = f2[x] = -(5/ [-x] ) +6 = -(5/ x ) +6

let g[x] = 5/(x-6)

x reflection: -g[x] = f1[x] = -[ 5/(x-6) ] = -5/(x-6)

y reflection: g1[-x] = f2[x] = -5/( [-x] - 6) = -5/( -x - 6)

hope that helps!

PizzaMaster

Hi everyone,

Is (2x-1) the same as (x-1/2)
I mean they both give x= 1/2
But let's say if we had to factorise 12x³ + 20x² − x − 6
and we get (2x-1)(2x+3)(3x+2)

Can I also write this as:
(x-1/2)(2x+3)(3x+2)?

ALSO can some please help and find the roots of this: 12x³ + 20x² − x − 6
using rational root theorem

much appreciated!!!

crumblycupcakes

PizzaMaster
good question! to answer your first one, (2x-1) = 2 (x-1/2). they're not the same but the same x value will make both of them equal (0). I'm not home rn so I'll reply to the second one when I have access to my iPad.

EDIT
I think it's easier if you watch a video, here's a good one I found (watching 0:00-2:40 should be enough): https://www.youtube.com/watch?v=4XytYH35AP0

PizzaMaster

So I can write (2x-1)(2x+3)(3x+2) as 2(x-1/2)(2x+3)(3x+2) NOT (x-1/2)(2x+3)(3x+2)
Great, thanks crumblycupcakes!

With rational root theorem, I find that it is super time consuming... after we find the possible roots, is there a shortcut to know which ones to try first?
For example with 12x³ + 20x² − x − 6:
the possible roots are +- [1/2, 1/3, 1/4, 1/6, 1/12, 2/3, 3/4, 3/2]
That is a lot! Is there a way to know which one we can try first?

Also does simplifying work?
for example with 12x³ + 20x² − x − 6, we do like factors of 6/ factors of 12. Can we do factors of 1/ factors of 2 (since 6/12 = 1/2?)

chemistry1111

show that 0=x² + 8x +17 has no solutions

d= (8)² -4(1)(17) =-4 is this the working out

crumblycupcakes

chemistry1111

Yep I think that should be enough but just to be safe and make sure I 💯 got the mark, I do something like

⁂ d<0 --> no real solutions

Billzene

crumblycupcakes yeah you never know if VCAA will reject "no solutions" on the grounds that there are complex solutions

crumblycupcakes

Billzene
I think that would only be the case in Specialist maths with VCAA so 'no solutions' is usually accepted in methods because complex numbers aren't taught in it. BUT we can't say for sure so it's always good to say 'no real solutions'. 😅

[ Edit ]
istg VCAA is so weird. They changed the study design for spesh and they included proofs but didn't specify if it was matrix proofs, number proofs or circle proofs or all of them. Apparently my teacher emailed them a couple of times but they never replied.

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