VCE Methods Question Thread!

[huehue]

Yep, Power Functions are in the new study deisgn.

I will try to do it, no guarantees if its right.

From the Question given, you know you need to solve it in the sequence T(X + B) = X'

Which brings you to:

-1(x-1)=x'
-2(y-5)=y'

Rearranging gives you:

x=-x'+1
y=-2y'+5

Should be easy from here, just sub in

-2y'+5= I -x'+1+2 I -3
-2y= I -x'+3 I -8
y= (I -x'+3 I)/-2 +4

Sorry have no clue how to use LATEX, need to learn 

thanks for the help but shouldn't rearranging give you x=-x'+1 and y=-y'/2 +5? unless i did something wrong lol

[Syndicate]

thanks for the help but shouldn't rearranging give you x=-x'+1 and y=-y'/2 +5? unless i did something wrong lol

No, you are right

[qazser]

thanks for the help but shouldn't rearranging give you x=-x'+1 and y=-y'/2 +5? unless i did something wrong lol

theres a minus in front of x, whoops can't divide  the y bit 

*facepalm* 

y=-2 I-x'+3I+4


i think thats right 

[huehue]

theres a minus in front of x, whoops can't divide  the y bit 

*facepalm* 

y=-2 I-x'+3I+4


i think thats right 

wait what, i got y= 2 |x-3| + 16 as my answer can you guys see what I did wrong cause I can't. here's my working out after subbing x and y into the equation:

-y/2 + 5 = |-x+1+2|-3
-y+2= |-x +3| -8
-y= 2|-x+3|-16
-y= -2|x-3|-16
y= 2|x-3|+16

[qazser]

wait what, i got y= 2 |x-3| + 16 as my answer can you guys see what I did wrong cause I can't. here's my working out after subbing x and y into the equation:

-y/2 + 5 = |-x+1+2|-3
-y+2= |-x +3| -8
-y= 2|-x+3|-16
-y= -2|x-3|-16
y= 2|x-3|+16

wrong again, think i need to go back to year 7

[Syndicate]

wait what, i got y= 2 |x-3| + 16 as my answer can you guys see what I did wrong cause I can't. here's my working out after subbing x and y into the equation:

-y/2 + 5 = |-x+1+2|-3
-y+2= |-x +3| -8
-y= 2|-x+3|-16
-y= -2|x-3|-16
y= 2|x-3|+16

That is the corrrect answer

wrong again, think i need to go back to year 7

LOL, it sometimes happens with me too, did it in my Year 10 exam  (silly mistakes can cost you a lost!)

[qazser]

Where's everyone up to for Methods 3/4 ? I've been lazy the past 2 weeks, starting 4C atm.

[William3558]

Hi,
Bit of a silly question. (√ is supposed to be root) I understand that 1/√2 is equal to √2/2. As you can multiply 1/√2 by √2/√2 to get √2/2.  Although I am not sure how to reverse it. Bit of an odd question but it is bugging me.
Cheers,

[Syndicate]

Hi,
Bit of a silly question. (√ is supposed to be root) I understand that 1/√2 is equal to √2/2. As you can multiply 1/√2 by √2/√2 to get √2/2.  Although I am not sure how to reverse it. Bit of an odd question but it is bugging me.
Cheers,

divide by , which basically means multipying it.

===>

===> =

If you dont know the dividing rules, well it basically means that you would flip the denominator and the numerator

example:

= =

[Phy124]

example:

= =

[qazser]

Phy out of nowhere spots a mistake!

i think he meant root 2 over 2

[Syndicate]

Thanks for noticing! I was copying and pasting it to save time. I will edit it now

[Phy124]

No worries at all

While I'm here I might as well offer you some LaTeX pointers:

Rather than typing ===> you can use "\Rightarrow" to produce and similarly if you wanted to have <=> you can use "\Leftrightarrow" to produce

Also instead of having

Code: [Select]

[tex]\frac { 2}{2 \sqrt 2} [/tex] = [tex]\frac {1}{\sqrt 2}[/tex]
You can simply have

Code: [Select]

[tex]\frac { 2}{2 \sqrt 2} = \frac {1}{\sqrt 2}[/tex]
Lastly something that might come in handy for the future. If you want to have something like



Start with "\begin{align}", start new lines with "\\" and place ampersand's before what you want to align before finishing with a "\end{align}", like so:

Code: [Select]

\begin{align}
x^2+5x+6 &= x^2 + 3x + 2x + 6
\\&=x(x+3)+2(x+3)
\\&=(x+2)(x+3)
\end{align}

Enjoy!

=

Also some people (for the most-part I personally couldn't care tbh) are always keen on rationalising the denominator so don't forget you can always use the nifty trick you mentioned earlier to rewrite that as if you wish

[Springyboy]

Im getting a bit jammed on this one so can I have some step-by-step instructions on how to do it
For h(x)=81x^4-72x^2+16, find the coordinates of the points where the graph of y=h(x) intersects the x- and y-axes and hence sketch the graph
It does say hint: first express h(x) as the square of a quadratic expression but I'm still getting stuck on it.
Sketching is fine and finding the y-int is fine, but how do I deal with factorising it?

[qazser]

Im getting a bit jammed on this one so can I have some step-by-step instructions on how to do it
For h(x)=81x^4-72x^2+16, find the coordinates of the points where the graph of y=h(x) intersects the x- and y-axes and hence sketch the graph
It does say hint: first express h(x) as the square of a quadratic expression but I'm still getting stuck on it.
Sketching is fine and finding the y-int is fine, but how do I deal with factorising it?

Look at it carefully, its the difference of 2 quartics, similar to difference of 2 squares.

Your answer should result in (x-a)^2(x+a)^2

Look at this carefully. You know it is a quartic

Find what to the power of 4 gives you 81 and what gives you 16

You should have got 3 and 2 respectively

Therefore the equation is (3x-2)^2(3x+2)^2