VCE Methods Question Thread!

[Jason12]

Do you mean ?

The expression simplifies to , which is just .

yeah I meant that

[alchemy]

The function with the rule y=x^4-x^3-10x^2+4x+24 is translated 2 units right and 5 units up. For the graph of the transformed function, the number of x-intercepts after translation is?

[brightsky]

y = x^4-x^3-10x^2+4x+24 = (x-3)(x-2)(x+2)^2

the transformed function is:

y = x^2 (x-4) (x-5) + 5

now let's investigate the function y = x^2 (x-4)(x-5). we know that the graph of this function is a positive quartic. we can also deduce that the function has two local minimums and one local maximum, and has the shape of a W. one local minimum, we know, is (0,0). the other local minimum occurs between x = 4 and x = 5. notice that if the y-value of this local minimum is greater than -5, then the transformed function would have no x-intercepts. if the y-value of this local minimum is less than -5, then the transformed function would have 2 x-intercepts. if the y-value of this local minimum is exactly equal to 5, then the transformed function will have one x-intercept. in order to find out the y-value of this local minimum, we shall derive the transformed function:

dy/dx = x(4x^2 - 27x + 40) = 0

we are only interested in the solution between 4 and 5 so:

x = (27+sqrt(89))/8

if we plug this into our original transformed function, we get:

y = (4933 - 801sqrt(89))/512 < -5

hence there are two x-intercepts.

[Jason12]

Can someone explain this rule
f(x)=x^1/3
f'(x) =1/3x^-2/3
f'(27)=1/3 * 1/(27^1/3)^2

And also negative powers like 2^-2 and 5^-3

[revcose]

If you didn't know, f'(x) =
To differentiate , you use the rule


Then they are simply replacing x with 27, but the reason it has changed so much is because of the index laws. The relevant ones for what you're wondering are:

When you have a negative power, you can remove the negative and switch what's left between numerator and denominator.  Generally you want to do this and avoid negative indices.

In your case, m = 2 and n = 1/3, so x^2/3 is the same as (x^1/3)^2

[ETTH96]

What are the transformations:

from f(x)= x^2     to    f(x)= 2(x-3)^2 +13

I know its
translation right 3 units
translation up 13 units
but what is the dilation?

I get really confused with dilation.. whether its dilation about the x or y, and if its 2 or 1/2..

or am i completely wrong?! I'm so confused... does the translation become right 6 units? what!??!?!

[rhinwarr]

Since the dilation is out of the brackets you don't need to worry about changing the translations.
Eg. If it was f(x)=(2x+3)^2
The horizontal translation would be 3/2 units left.

In the case of parabolas, dilation by a factor of 2 from x axis is the same as dilation by factor of 1/2 from the y axis so both versions are correct. However if it is a log or exponential graph it is different so watch out for those.

Generally if the dilation factor is out of the brackets I call it a dilation from the x axis. If the dilation is in the form of the coefficient of x (eg. Log(2x+3) )  then it's dilation from the y axis

[lzxnl]

What are the transformations:

from f(x)= x^2     to    f(x)= 2(x-3)^2 +13

I know its
translation right 3 units
translation up 13 units
but what is the dilation?

I get really confused with dilation.. whether its dilation about the x or y, and if its 2 or 1/2..

or am i completely wrong?! I'm so confused... does the translation become right 6 units? what!??!?!

I personally use the dash method of transformations to help me read off what the transformations are.
Firstly, you compare the two equations. In the transformed equation, you replace x with x' and y with y'.
So, we have y=x^2 and y'=2(x-1)^2+13
We need to rearrange the right hand side in the form a(y'-b)=(x'-d)^2 or something similar, so after rearranging, we get (y'-13)/2=(x'-1)^2

Compare this to our original equation. By directly comparing y's and x's, we see that y=(y'-13)/2 if comparing the left hand sides and x'-1=x
Solving for the dashed variables, we get y'=2y+13 and x'=x+1
So how do we interpret this? We read these transformations in the order of BODMAS. So for x, we just have a translation of one unit in the positive x direction as the transformed variable is one larger than the original variable.
For y, we firstly have a dilation factor two from the x axis and THEN a translation 13 units in the positive y direction. If mastered, this method is an almost fool-proof method of working out transformations. Don't bother trying to remember.

As for dilations, think of it this way. A dilation from the x axis doesn't actually affect the horizontal x variable. Likewise, a dilation from the y axis doesn't affect the vertical y variable.

[Only Cheating Yourself]

Guys, i had a look over the year 9,10 textbook i can do most of the chapters, however there are a few i can't for e.g find the % of…. What should i do?

[ShortBlackChick]

Go through those chapters maybe?

[Only Cheating Yourself]

Go through those chapters maybe?

but if there not going to be relevant then i shouldn't should i?  I've went over what i should know our school gave a booklet which consisted of linear equations, graphs, simultaneous equations etc, I'm fine with that…

[bucklr]

but if there not going to be relevant then i shouldn't should i?  I've went over what i should know our school gave a booklet which consisted of linear equations, graphs, simultaneous equations etc, I'm fine with that…

It can't hurt to lightly touch on the chapters you don't know how to do right? Especially if it gives you peace of mind. In all honesty the chapters from year 9 and 10 aren't going to be of great relevance. As you go through the course during the year you can always ask your teacher to give you a refresher - so don't stress, just do your best.

[smile+energy]

I need some help
1.What is the best way to solve it? Q: linear simultaneous equations: mx-5y=10 and 3x-(m-2)y=6.
   a)Find the values of m, where m belongs to R, for which are infinitely many or no solutions.
   b)Find the unique solution for the equation in terms of m.

2. Do I need to learn how to solve the questions algebraically? My teacher said we don’t need to know it.
X+y+z=9
-x+2y-3z+-15
X+5y+3z=29
Thanks in advance.

[Stick]

1. Do you know how to solve simultaneous equations using matrices? If you can but can't remember, here's a hint: it involves working out the determinant.

2. Not really, most of the time you get that you will have your calculator to help you. However, it isn't a bad idea to try it by hand.

[Only Cheating Yourself]

Ian is six years older than Liz and their combined ages total sixty years.  How old is each person

let Liz=x and Ian=6+x
2x+6=60
2x=54
x=27
therefore ian is 33

This is wrong, could someone help me please?