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May 11, 2021, 07:44:23 am

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#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18810 on: September 23, 2020, 04:00:47 pm »
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you know what, **** it. first I had failed security checks then it said i already posted, but i didn't, and then it said IP address try again later.

here's the question copy and pasted from elsewhere.

The graph of the function y=x^4−12x^3 is translated by a units in the positive direction of the x-axis and b units in the positive direction of the y-axis (where a and b are positive constants).
a) Find the coordinates of the turning points of the graph of y=x^4−12x^3.
b) Find the coordinates of the turning points of its image.

specifically b.

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18811 on: September 23, 2020, 04:42:02 pm »
0
you know what, **** it. first I had failed security checks then it said i already posted, but i didn't, and then it said IP address try again later.

here's the question copy and pasted from elsewhere.

The graph of the function y=x^4−12x^3 is translated by a units in the positive direction of the x-axis and b units in the positive direction of the y-axis (where a and b are positive constants).
a) Find the coordinates of the turning points of the graph of y=x^4−12x^3.
b) Find the coordinates of the turning points of its image.

specifically b.
you know what, **** it. first I had failed security checks then it said i already posted, but i didn't, and then it said IP address try again later.

here's the question copy and pasted from elsewhere.

The graph of the function y=x^4−12x^3 is translated by a units in the positive direction of the x-axis and b units in the positive direction of the y-axis (where a and b are positive constants).
a) Find the coordinates of the turning points of the graph of y=x^4−12x^3.
b) Find the coordinates of the turning points of its image.

specifically b.

Sure - what have you tried?

#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18812 on: September 23, 2020, 04:44:09 pm »
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Sure - what have you tried?
I get (-a,b), whereas they get (a,b). They get (9+a, -2187+b), whereas I get (9-a,-2817+b)

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18813 on: September 23, 2020, 04:49:38 pm »
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I get (-a,b), whereas they get (a,b). They get (9+a, -2187+b), whereas I get (9-a,-2817+b)

Not what I asked - what method did you do to get to this point?

#### cinnamonscroll

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##### Re: VCE Methods Question Thread!
« Reply #18814 on: September 23, 2020, 05:02:34 pm »
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Hi!

I was wondering if anyone has a list of irrelevant probability exam questions because I find it hard to tell which questions I should/shouldn't do lol

Thanks~

#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18815 on: September 23, 2020, 05:12:35 pm »
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Not what I asked - what method did you do to get to this point?
I took the derivative of the given function, found the turning points of that, then determined the turning points.

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18816 on: September 23, 2020, 05:28:38 pm »
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I took the derivative of the given function, found the turning points of that, then determined the turning points.

Okay, so I'm guessing there may a confusion then in how you apply the transformations on a single point. So, what you should've gotten from differentiating is that the turning points of the original function are at:

(0, 0), (9, -2187)

We're translating x in the positive direction by a units, and y in the positive direction by b units. So, this means we're taking on the mapping:

(x', y') --> (x+a, y+b)

Where x' and y' are the images of x and y. So you see, all we need to do to get to those images, is add a and b:

(0+a,0+b), and (9+a,-2187+b)

So, I think the confusion lies in the fact that you probably know to translate an equation in the positive direction, we'd normally do so like this:

f(x-a)+b

The key point here, is that this is for an equation mapping, NOT a point mapping. Here's why:

You can think of our mapping as one in which you have two new equations:

x' = ax+b
y' = cy+d

Now, if I want to put these into the equation y=f(x), I need to solve them for x and y - I can't just substitute x' and y' directly into the equation, because the equation isn't in terms of them - it's in terms of x and y. So, I solve them both:

(x'-b)/a = x
(y'-d)/c = y

I put them into my equation y=f(x):

$y=f(x)\\
\frac{y'-c}{d}=f\left(\frac{x'-b}{a}\right)\\
y'=d\times f\left(\frac{x'-b}{a}\right)+c$

Which is why for the equations, we minus the translation on x, but for single point mappings, we add the translation on x instead.

Hi!

I was wondering if anyone has a list of irrelevant probability exam questions because I find it hard to tell which questions I should/shouldn't do lol

Thanks~

I don't, sorry, but essentially you can think of it like this - if the probability question looks like something you could answer in year 11, it's probably in the study design still. If it's about binomial distributions or normal distributions, it's in the study design still. If it doesn't fall into one of those three categories, it's probably not in the study design still.

#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18817 on: September 25, 2020, 09:17:20 pm »
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Hello!

I have a question
A piece of string 10 metres long is cut into two pieces to form two squares.
a) If one piece of string has length x metres, show that the combined area of the two
squares is given by A = 1/8 (x^2 − 10x + 50).

This seems like a pretty basic question, but I dunno where I went wronggg ( help would be rlly appreciated!
Just someone who likes to learn a lot of questions.

Aspiring medical practitioner! ʕ•́ᴥ•̀ʔっ

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18818 on: September 25, 2020, 09:18:38 pm »
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Hello!

I have a question
A piece of string 10 metres long is cut into two pieces to form two squares.
a) If one piece of string has length x metres, show that the combined area of the two
squares is given by A = 1/8 (x^2 − 10x + 50).

This seems like a pretty basic question, but I dunno where I went wronggg ( help would be rlly appreciated!

Interesting! Why don't you tell us what you tried, then we can tell you where you went wrong

#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18819 on: September 26, 2020, 10:31:30 am »
-1
Hello!

I have a question
A piece of string 10 metres long is cut into two pieces to form two squares.
a) If one piece of string has length x metres, show that the combined area of the two
squares is given by A = 1/8 (x^2 − 10x + 50).

This seems like a pretty basic question, but I dunno where I went wronggg ( help would be rlly appreciated!

Is this 3/4 or 1/2? I remember doing this not long ago.

Let x = length of string for one square and y = length of string for second square.

Then:

x + y = 10
y = 10 - x

The length of one side of the square with the x string is x/4. The length of a side of the square with the y string is y/4.

You should be able to work out the rest.

#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18820 on: September 26, 2020, 10:41:34 am »
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Interesting! Why don't you tell us what you tried, then we can tell you where you went wrong
Is this 3/4 or 1/2? I remember doing this not long ago.

Let x = length of string for one square and y = length of string for second square.

Then:

x + y = 10
y = 10 - x

The length of one side of the square with the x string is x/4. The length of a side of the square with the y string is y/4.

You should be able to work out the rest.

Ohhhhhhh! I got it now, I actually had x as the side length of one of the squares :0, so I got the answer 2x^2 - 5x + 25/4. Thank you guys!
Just someone who likes to learn a lot of questions.

Aspiring medical practitioner! ʕ•́ᴥ•̀ʔっ

#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18821 on: September 26, 2020, 11:37:00 am »
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Ohhhhhhh! I got it now, I actually had x as the side length of one of the squares :0, so I got the answer 2x^2 - 5x + 25/4. Thank you guys!
haha i made the same error when i first did the question.

#### thatdumbstudent

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##### Re: VCE Methods Question Thread!
« Reply #18822 on: September 26, 2020, 07:57:39 pm »
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hi guys, could someone please explain the answer to this transformation question for me? (image attached)

so the answer is E, and i've looked at the worked solutions (also attached) and i get every single line of working up until the "Now..." bit? like where does that equation come from? is it just from simplifying, if so, could someone please explain to me how it works? tia!!

#### The Cat In The Hat

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##### Re: VCE Methods Question Thread!
« Reply #18823 on: September 26, 2020, 09:19:05 pm »
+2
hi guys, could someone please explain the answer to this transformation question for me? (image attached)

so the answer is E, and i've looked at the worked solutions (also attached) and i get every single line of working up until the "Now..." bit? like where does that equation come from? is it just from simplifying, if so, could someone please explain to me how it works? tia!!
The 'now' bit appears to just be rearranging and simplifying. Try rearranging it until y' is all you have on the LHS, and see if you can do it from there.
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#### thatdumbstudent

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##### Re: VCE Methods Question Thread!
« Reply #18824 on: September 26, 2020, 09:46:23 pm »
+1
The 'now' bit appears to just be rearranging and simplifying. Try rearranging it until y' is all you have on the LHS, and see if you can do it from there.

got it now! turns out i wasn't reading the question properly and ignored the g(x) given,, thank you!