VCE Methods Question Thread!

[Yertle the Turtle]

When you said the work required will be different for everyone, is there still like a number of hours that we should do methods each week, and number of practise exams for an average student to get 35 raw?

People work in different ways. For some people to understand a concept they will need to go over and over and over it again, while for others (me, to some extent) it is much more intuitive. In general just find what works for you, and just put in your best effort.

[Sine]

When you said the work required will be different for everyone, is there still like a number of hours that we should do methods each week, and number of practise exams for an average student to get 35 raw?

there is no formula of X hours = Y study score just put in the time and effort that you can throughout the year, find your weaknesses - correct them

[snowisawesome]

Thanks sine and Marvin K Mooney
Have another question
State the sequence that takes y = x^2 to y = -2(1-x)^3+1
The answered mentioned that y = x^3 has been translated 1 unit to the right and reflected in the x and y axis
Could someone please explain why these occur?
Thanks

[True Dat]

Thanks sine and Marvin K Mooney
Have another question
State the sequence that takes y = x^2 to y = -2(1-x)^3+1
The answered mentioned that y = x^3 has been translated 1 unit to the right and reflected in the x and y axis
Could someone please explain why these occur?
Thanks

How does a transformation turn a quadratic into a cubic???
As for the reflection over the x axis , that is because of the (-) in front of the 2, and the reflection over the y axis is because of the (-) in front of the x.
EDIT: What Sine said is correct, all hail Sine D:
Hope this answered your query

[snowisawesome]

How does a translation turn a quadratic into a cubic???
As for the reflection over the x axis , that is because of the (-) in front of the 2, and the reflection over the y axis is because of the (-) in front of the x. In regards to the 1 unit to the right, I think it should be 1 to the left because -(x)+1 is equivalent to a translation of 1 unit to the left while -(x)-1 is equivalent to a translation of 1 unit to the right. And (1-x) is equivalent to -x+1.
Hope this answered your query

Sorry accidentally wrote x^2 instead of x^3

[Sine]

Thanks sine and Marvin K Mooney
Have another question
State the sequence that takes y = x^2 to y = -2(1-x)^3+1
The answered mentioned that y = x^3 has been translated 1 unit to the right and reflected in the x and y axis
Could someone please explain why these occur?
Thanks

y = x^2 to y = -2(1-x)^2+1

let's put the funciton in an easier to read form first
y = -2(-(x-1))^2 + 1
y = -2(x-1)^2 +1

Dilation by factor 2 from the x-axis (swap 'y" with "y/2")
y/2 = x^2
y = 2x^2

Reflect in the x-axis (swap "y" with "-y")
-y = 2x^2
y = -2x^2
Translation 1 unit in the positive direction of the x-axis (swap "x" with "x-1")
y = -2(x-1)^2
Translation 1 unit in the positive direction of the y-axis (swap "y" with "y-1")
y - 1 = -2(x-1)^2
y = -2(x-1)^2  + 1
y = -2(1-x)^2 + 1

[lzxnl]

This is how I do things. Compare

to


Rearrange the second equation to get

Clearly, it suffices to take

y' is the new variable. How do we get the new y value from the old? Simple. As y' = 1 - 2y, the first operation on the y (to get to the new one) is doubling, i.e. dilation factor 2 from x axis. Then, the next operation is the negative, so reflection in x axis. Then, we have a +1, which is a translation by 1 unit upwards.
To get from the old x value to the new one, we need to multiply by -1, aka reflect in the y axis, then translate right by 1 because of the +1. This method makes transformations pretty mechanical and straightforward.

mod edit: fixed latex code (wasn't displaying properly)

[snowisawesome]

A few questions
If y = 0.5*((x+4)/2)^2-1
then is it also possible to write it as
0.5*((0.5*(x+4))^2-1)

If -y-2 = cos(x-(pi/4))
Then -y = cos(x-(pi/4))+2
then wouldn't y = - (cos(x-(pi/4))+2)
which is y = -cos(x+(pi/4)-2
but the answer said it's y = -cos(x-(pi/4)-2

The original function is y = x^3
it becomes -y = ((0.5)(x+4))^3
it said that y = x^3 was dilated by a factor of 2 from the y axis, reflected in the x axis and  translated 4 units to the left
But the answer also said that the transformed rule was y = -0.125(x+4)^3
Can someone please explain why it was dilated by a factor of 2 from the y axis when it says -0.125, which means dilated by a factor of 8 from the y axis?

Thanks

Also,
y = -2(1-x)^3 + 1
The answered mentioned that y = x^3 has been translated 1 unit to the right and reflected in the x and y axis
Could someone please explain why these occur? I'm especially confused when it said translated 1 unit to the right when it says
y = -2(1-x)^3 + 1 when it said 1-x it means -x+1, doesn't that mean translated 1 unit to the left because of the +1?
And could someone also explain how they get the reflection in the x and y axis?
I know i've asked this before, but i'm still confused with it
thanks again

[Yertle the Turtle]

Also,
y = -2(1-x)^3 + 1
The answered mentioned that y = x^3 has been translated 1 unit to the right and reflected in the x and y axis
Could someone please explain why these occur? I'm especially confused when it said translated 1 unit to the right when it says
y = -2(1-x)^3 + 1 when it said 1-x it means -x+1, doesn't that mean translated 1 unit to the left because of the +1?
And could someone also explain how they get the reflection in the x and y axis?
I know i've asked this before, but i'm still confused with it
thanks again

-2(1-x)^3 becomes -2(-x+1)^3, but in t.p. form there has to be a positive x with a coefficient of 1, so you would take the - out the front of the brackets: 2(x-1)^3

[snowisawesome]

-2(1-x)^3 becomes -2(-x+1)^3, but in t.p. form there has to be a positive x with a coefficient of 1, so you would take the - out the front of the brackets: 2(x-1)^3

Thanks Marvin
Also, is it possible for you to answer my other questions?
If y = 0.5*((x+4)/2)^2-1
then is it also possible to write it as
0.5*((0.5*(x+4))^2-1)

If -y-2 = cos(x-(pi/4))
Then -y = cos(x-(pi/4))+2
then wouldn't y = - (cos(x-(pi/4))+2)
which is y = -cos(x+(pi/4)-2
but the answer said it's y = -cos(x-(pi/4)-2

The original function is y = x^3
it becomes -y = ((0.5)(x+4))^3
it said that y = x^3 was dilated by a factor of 2 from the y axis, reflected in the x axis and  translated 4 units to the left
But the answer also said that the transformed rule was y = -0.125(x+4)^3
Can you please explain why it was dilated by a factor of 2 from the y axis when it says -0.125, which means dilated by a factor of 8 from the y axis?

State the transformations that have been applied to y = 1/x to get y = 2-3/(x-1)
The answer said that one of the transformations that took y = 1/x to get y = 2-3/(x-1) was that y = 1/x has been reflected in the x axis or the y axis
Could you please explain why this is so?

Thanks

[Bell9565]

Thanks Marvin
Also, is it possible for you to answer my other questions?
If y = 0.5*((x+4)/2)^2-1
then is it also possible to write it as
0.5*((0.5*(x+4))^2-1)

Look I'm not Marvin but I can help with these if you would like

0.5*((x+4)/2)^2-1 does not equal 0.5*((0.5*(x+4))^2-1)
the 0.5 which is multiplied against the brackets is only against the brackets, thus making it one term whilst the -1 is a separate term
aditionally the /2 inside the bracket is also being squared therefore it cannot be factorised out

If -y-2 = cos(x-(pi/4))
Then -y = cos(x-(pi/4))+2
then wouldn't y = - (cos(x-(pi/4))+2)
which is y = -cos(x+(pi/4)-2
but the answer said it's y = -cos(x-(pi/4)-2

When you apply the negative to the y = - (cos(x-(pi/4))+2), the x-(pi/4) bit is part of what the cos is doing so you don't apply the negative from the outside to it. I'm not sure how to explain what I mean but because the term is (os(x-(pi/4) you don't change the sign.


I'm not great with transformations so I can't really help with the last 2, I'm sure someone else will!

Hope I helped a little!

[Sine]

The original function is y = x^3
it becomes -y = ((0.5)(x+4))^3
it said that y = x^3 was dilated by a factor of 2 from the y axis, reflected in the x axis and  translated 4 units to the left
But the answer also said that the transformed rule was y = -0.125(x+4)^3
Can you please explain why it was dilated by a factor of 2 from the y axis when it says -0.125, which means dilated by a factor of 8 from the y axis?

State the transformations that have been applied to y = 1/x to get y = 2-3/(x-1)
The answer said that one of the transformations that took y = 1/x to get y = 2-3/(x-1) was that y = 1/x has been reflected in the x axis or the y axis
Could you please explain why this is so?

Thanks

for the first one (following the books answer)
y=x^3
dilation factor 2 from y axis
y = (x/2)^3
y = x^3/8
reflect in x-axis
-y = x^3/8
y = -x^3/8
translate 4 units to the left
y = -0.125(x+4)^3

2nd q

y = 1/x to get y = 2-3/(x-1)

let's get the final function in a better form
(y-2)/-3 = 1/(x-1)
So it goes from y=1/x
Dilation of scalar factor 3 from the x-axis
y/3 = 1/x
y=3/x
Reflection in the x-axis
-y = 3/x
y = -3/x
Translation 2 units in the positive direction of the y-axis
y-2 = -3/x
y = -3/x + 2
Translation 1 unit in the positive direction of the x-axis
y = -3/(x-1) + 2

[Yertle the Turtle]

The original function is y = x^3
it becomes -y = ((0.5)(x+4))^3
it said that y = x^3 was dilated by a factor of 2 from the y axis, reflected in the x axis and  translated 4 units to the left
But the answer also said that the transformed rule was y = -0.125(x+4)^3
Can you please explain why it was dilated by a factor of 2 from the y axis when it says -0.125, which means dilated by a factor of 8 from the y axis?

State the transformations that have been applied to y = 1/x to get y = 2-3/(x-1)
The answer said that one of the transformations that took y = 1/x to get y = 2-3/(x-1) was that y = 1/x has been reflected in the x axis or the y axis
Could you please explain why this is so?

Thanks

Not sure about the first one, it doesn't really make sense to me. However the second one does make sense.

y=2-(3/(x-1)) is the same as y=-(3/(x-1))+2
Therefore the following transformations apply:
dilation of 3 from the x-axis
reflection in the x/y-axis
translation of 1 in the x-axis and 2 in the y-axis

The reason you can reflect in either the x or the y-axis is because this graph is an untranslated rectangular hyperbola, and therefore the reflections will be the same whichever way you do it.

Hope this helps!

[snowisawesome]

for the first one (following the books answer)
y=x^3
dilation factor 2 from y axis
y = (x/2)^3
y = x^3/8
reflect in x-axis
-y = x^3/8
y = -x^3/8
translate 4 units to the left
y = -0.125(x+4)^3

2nd q

y = 1/x to get y = 2-3/(x-1)

let's get the final function in a better form
(y-2)/-3 = 1/(x-1)
So it goes from y=1/x
Dilation of scalar factor 3 from the x-axis
y/3 = 1/x
y=3/x
Reflection in the x-axis
-y = 3/x
y = -3/x
Translation 2 units in the positive direction of the y-axis
y-2 = -3/x
y = -3/x + 2
Translation 1 unit in the positive direction of the x-axis
y = -3/(x-1) + 2

How did you go from y = -x^3/8 to y = -0.125(x+4)^3

Also, do you know how 0.5*((x+4)/2)^2-1 can be written in another form
Thanks

[Sine]

How did you go from y = -x^3/8 to y = -0.125(x+4)^3



Also, do you know how 0.5*((x+4)/2)^2-1 can be written in another form
Thanks

y = -x^3/8 is basically y = -x^3 * 1/8

1/8 = 0.125

0.5*((x+4)/2)^2-1

maybe in a fractional form which is much easier to read so
1/2 * (x+4)^2/4 -1
(x+4)^2/8 -1