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Math question

Can someone please help with this?

(a)

i) 1/3 y= f(x) --> y=3f(x) I get this

ii) I don’t understand this

iii) y= f(2x) I get this

iv) I don’t understand this

v) Shouldn’t it be y=3f(2x (a+b/2))+k ?


(b) I have no clue :frown:

I have attached the question and mark scheme below.

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(edited 6 years ago)
you need to move the graph to the left, which means that the thing goes inside the bracket ( f(x+something) ). You want to move it to the left by the middle value, which can be determined by the average of both p and q, so you do (0.5x(p+q)). Therefore the answer is f(x+0.5(p+q)).

This is the answer to ii.
Reply 2
Original post by WillyWonka96
you need to move the graph to the left, which means that the thing goes inside the bracket ( f(x+something) ). You want to move it to the left by the middle value, which can be determined by the average of both p and q, so you do (0.5x(p+q)). Therefore the answer is f(x+0.5(p+q)).

This is the answer to ii.


Thanks but the answer to ii) is y=f(x - ((a+b)/2)). This is shown in the mark scheme which I have attached in my first post.
(a+b)/2 is the same as (p+q)/2. Since they are equidistant it doesn't make a difference. Same mathematical principles apply. Hope this helps
for iv, putting the '+k' at the end of f(x) just adjusts the graph in the y direction, which the question asks for
Original post by sienna2266
...


(ii) I certainly had to take a look at the answers to figure out what they even mean by part (ii) but once you're happy that the middle of the 'dip' occurs half-way between t=at=a and t=bt=b, then the translation should be obvious as you want this bit to occur at t=0t=0. As this is a periodic function, it doesn't matter whether you do f(t+a+b2)f(t+\frac{a+b}{2}) or f(ta+b2)f(t-\frac{a+b}{2})

(iv) The amplitude is yy, so increasing it by kk means moving it up by kk hence y(yk)y \mapsto (y-k)

(v) Yes it would be what you say

(b) Just sketch 4 different graphs, applying transformation after transformation, and you'll get to the last one.

EDIT: I just realised their graph axes don't make sense either. It should be f(t)f(t) on the vertical, and y=f(t)y=f(t) in the context.
(edited 6 years ago)
Reply 6
Original post by sienna2266
Thanks but the answer to ii) is y=f(x - ((a+b)/2)). This is shown in the mark scheme which I have attached in my first post.


Thanks but I am really confused .. So what is ii) asking for? What does zero time mean? and is the dip the distance between p and q?
Original post by sienna2266
Thanks but I am really confused .. So what is ii) asking for? What does zero time mean? and is the dip the distance between p and q?


zero time is when x=o
Reply 8
Original post by RDKGames
(ii) I certainly had to take a look at the answers to figure out what they even mean by part (ii) but once you're happy that the middle of the 'dip' occurs half-way between t=at=a and t=bt=b, then the translation should be obvious as you want this bit to occur at t=0t=0. As this is a periodic function, it doesn't matter whether you do f(t+a+b2)f(t+\frac{a+b}{2}) or f(ta+b2)f(t-\frac{a+b}{2})

(iv) The amplitude is yy, so increasing it by kk means moving it up by kk hence y(yk)y \mapsto (y-k)

(v) Yes it would be what you say

(b) Just sketch 4 different graphs, applying transformation after transformation, and you'll get to the last one.

EDIT: I just realised their graph axes don't make sense either. It should be f(t)f(t) on the vertical, and y=f(t)y=f(t) in the context.



Thank you so much -this is starting to make much more sense.

iv) The amplitude is yy, so increasing it by kk means moving it up by kk hence y(yk)y \mapsto (y-k)

I am a bit confused here with the answer you have given because the mark scheme answer is y=f(x) +k.

I know the amplitude of the signal is the height of the signal because it says so in (i)

The mark scheme answer y=f(x) +k doesn’t make much sense to me either because doesn’t this just shift the signal graph up instead of increasing the height/peak of the signal by k?

Also, is the mark scheme answer wrong for (v) ?
Reply 9
Original post by RDKGames
(ii) I certainly had to take a look at the answers to figure out what they even mean by part (ii) but once you're happy that the middle of the 'dip' occurs half-way between t=at=a and t=bt=b, then the translation should be obvious as you want this bit to occur at t=0t=0. As this is a periodic function, it doesn't matter whether you do f(t+a+b2)f(t+\frac{a+b}{2}) or f(ta+b2)f(t-\frac{a+b}{2})

(iv) The amplitude is yy, so increasing it by kk means moving it up by kk hence y(yk)y \mapsto (y-k)

(v) Yes it would be what you say

(b) Just sketch 4 different graphs, applying transformation after transformation, and you'll get to the last one.

EDIT: I just realised their graph axes don't make sense either. It should be f(t)f(t) on the vertical, and y=f(t)y=f(t) in the context.


Thank you so much -making much more sense.

iv) The amplitude is , so increasing it by means moving it up by hence

I am a bit confused here with the answer you have given because the mark scheme answer is y=f(x) +k.

I know the amplitude of the signal is the height of the signal because it says so in (i)

The mark scheme answer y=f(x) +k doesn’t make much sense to me either because doesn’t this just shift the signal graph up instead of increasing the height/peak of the signal by k?

Also, is the mark scheme answer wrong for (v) ?
Original post by sienna2266
Thank you so much -making much more sense.

iv) The amplitude is , so increasing it by means moving it up by hence

I am a bit confused here with the answer you have given because the mark scheme answer is y=f(x) +k.

I know the amplitude of the signal is the height of the signal because it says so in (i)

The mark scheme answer y=f(x) +k doesn’t make much sense to me either because doesn’t this just shift the signal graph up instead of increasing the height/peak of the signal by k?


But isn't that what shifting up the graph by kk does? Increase the peak by kk?

So then as I said, you replace yy by yky-k hence yk=f(t)y=f(t)+ky-k=f(t) \Rightarrow y=f(t)+k

Also, is the mark scheme answer wrong for (v) ?


I would say so.
Original post by RDKGames
But isn't that what shifting up the graph by kk does? Increase the peak by kk?

So then as I said, you replace yy by yky-k hence yk=f(t)y=f(t)+ky-k=f(t) \Rightarrow y=f(t)+k



I would say so.


Thanks very much.

So is the peak/amplitude of the graph measured from the x axis and not from the lowest point of the graph?
(edited 6 years ago)
Original post by sienna2266
Thanks very much.

So is the peak/amplitude measured from the x axis and not from the lowest point of the graph?


If you at the definitions: https://en.wikipedia.org/wiki/Amplitude I believe in this question they are using the Peak Amplitude rather than Peak-to-Peak, so they are measuring from the t-axis to the highest point.

TBH I'm not comfortable with this question and the way its worded/worked out.
Original post by RDKGames
If you at the definitions: https://en.wikipedia.org/wiki/Amplitude I believe in this question they are using the Peak Amplitude rather than Peak-to-Peak, so they are measuring from the t-axis to the highest point.

TBH I'm not comfortable with this question and the way its worded/worked out.


Thanks very much! :smile:

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