VCE Methods Question Thread!

[integrationbyrecognition]

methods nht exam 2 2018
mcq 14
for question 14, since it's a dilation of factor 1/2 from the x axis, don't you need to multiple the entire equation by 1/2, so shouldn't it be 3/2cos(-x-pi/3) -2?

[Lear]

methods nht exam 2 2018
mcq 14
for question 14, since it's a dilation of factor 1/2 from the x axis, don't you need to multiple the entire equation by 1/2, so shouldn't it be 3/2cos(-x-pi/3) -2?

Make sure you do the transformations in the order they are given.

[integrationbyrecognition]

Make sure you do the transformations in the order they are given.

So it you dilate first, then translate second, dilation won't affect the translation? I'm not really sure what you mean by order, could you please explain? And do you have an a+ cutoff prediction for yesterday's exam 1?
Thanks

[Lear]

Hey wouldnt the -y be a reflection in the x axis ? According to my notes it is in option C and D.

Note the 'fatness' of the transformed graph in comparison to the original one. The question states for a reason that both graphs 'have the same scale'

[Lear]

So it you dilate first, then translate second, dilation won't affect the translation? I'm not really sure what you mean by order, could you please explain? And do you have an a+ cutoff prediction for yesterday's exam 1?
Thanks

If you have dilated first then the dilations do not affect the translations. The order of transformation is vital and it is for this reason the questions states clearly '....in that order'

And no I don't have an A+ cut off prediction

[passbleh24]

Note the 'fatness' of the transformed graph in comparison to the original one. The question states for a reason that both graphs 'have the same scale'

ohhhh , Yes I do

Thank you!!

[integrationbyrecognition]

If you have dilated first then the dilations do not affect the translations. The order of transformation is vital and it is for this reason the questions states clearly '....in that order'

And no I don't have an A+ cut off prediction

Cheers, do you also think yesterday's exam 1 was harder than usual?

[Bri MT]

The linear part of the graph has an average value of 0.

Thank you!

I completely forgot to pay attention attention to  above/below the axis

[passbleh24]

Thank you!

I completely forgot to pay attention attention to  above/below the axis

How does this point change the answer from a to a/2 ? I'm still struggling to understand

Thanks

[MB_]

How does this point change the answer from a to a/2 ? I'm still struggling to understand

Thanks

The "area" under the linear part of the graph is 0. Adding that to a^2 and dividing by 2a gives a/2

[passbleh24]

The "area" under the linear part of the graph is 0. Adding that to a^2 and dividing by 2a gives a/2

What part of the graph are you specifically talking about ? I'm sorry its just not clicking

"Find the area by splitting into two shapes (triangle & square) rather than by integrating.
Area=.5a(2a) + a^2 = 2a^2
Then divide by a-(-a) which equals 2a
This gives you the answer of a"

^^ That is mini turtles working out and I did the exact same, do you mind pointing out specifically what section was wrong and why?

Thank for your assistance

[MB_]

What part of the graph are you specifically talking about ? I'm sorry its just not clicking

[Lear]

How would I find the average value over the interval -a to a ?

You can just find the equation of the lines in terms of a and integrate them over the given domain.

[passbleh24]

You can just find the equation of the lines in terms of a and integrate them over the given domain.

I think I might just do that.

(Image removed from quote.)

Thank you soo much !! I understand what you mean now.

[Freddie Hg]

what should i do before exam 2??