Login

Welcome, Guest. Please login or register.

March 28, 2024, 08:08:37 pm

Author Topic: Graphing question  (Read 3116 times)  Share 

0 Members and 1 Guest are viewing this topic.

mikamika

  • Adventurer
  • *
  • Posts: 6
  • Respect: 0
Graphing question
« on: October 21, 2018, 10:32:37 pm »
0
Hello!

I stumbled upon a question in the 2009 HSC paper (q3 aiii), and was wondering,
what is the difference between y=[f(x)^2] and y=(x^2)?

I have no idea how to draw y=(x^2) unless there's an equation or is downright obvious. In this case, don't know what they did at all...

Any help will be great

Cheers

3.14159265359

  • Trendsetter
  • **
  • Posts: 183
  • Respect: +16
Re: Graphing question
« Reply #1 on: October 21, 2018, 10:40:18 pm »
0
Hello!

I stumbled upon a question in the 2009 HSC paper (q3 aiii), and was wondering,
what is the difference between y=[f(x)^2] and y=(x^2)?

I have no idea how to draw y=(x^2) unless there's an equation or is downright obvious. In this case, don't know what they did at all...

Any help will be great

Cheers

hello

when the question says y=[f(x)]^2 that means its the ENTIRE FUNCTION THATS SQUARED so what you do is you square all the y values for a corresponding x value to obtain that graph. and obviously the graph will be above the x axis since the whole graph is squared. so if the original graph has point (2,5) your new graph will have point (2,25) or if its (1, -3) on new graph its (1,9)

for y=f(x^2) that means ALL THE X VALUES ARE SQAURED meaning that if the original graph has point (2,5) your new graph will have point (4,5) if its (-1,3) graph will be (1,3)

hope that makes sense and I did not make any errors explaining. if something is not clear ill try to explain further.

hope that helps :)

mikamika

  • Adventurer
  • *
  • Posts: 6
  • Respect: 0
Re: Graphing question
« Reply #2 on: October 21, 2018, 10:48:40 pm »
0
Hey,

thanks for your response

I'm still not too sure because if I apply your logic to this question, the answers don't match up (i've linked it for your convenience)

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Graphing question
« Reply #3 on: October 21, 2018, 11:12:37 pm »
+1
Their idea was on the right track, it's just that it worked backwards.

For \( y=f(x^2) \), you square the \(x\)-coordinate before applying the function to it. Since the square of any number is negative, effectively speaking anything that was to the left of the \(y\)-axis on the original graph gets discarded, because for something like \(f(x^2)\) you'll never end up plugging a number like -1 into \(f\).

You can see that the graph of \( y = f(x)\) passes through \( (0,0)\), so as you'd expect \(y=f(x^2)\) also passes through \( (0,0) \). Simple reason is because \(0^2 = 0\).

However, \(y=f(x)\) also passes through \( (4,0) \). Since \(4 = 2^2\), for \(y=f(x^2)\) you'd observe the point \( (2,0) \) to lie on there instead. Reason being if you plug \(x=2\) in, you get \(y = f(2^2)\), which becomes \(f(4)\) and equals to \(0\).
In a similar way, \( (-2,0) \) lies on the curve.

Noting that the pattern with the negatives always occurs, you also expect everything on the left of the \(y\)-axis to be a reflection of everything on the right. This can clearly be observed in their solution.

Also, quite visibly in their solution the graph looks more stretched inwards towards the \(y\)-axis. This can be thought of intuitively, as if \(x\) is increasing, \(x^2\) is increasing at a much faster rate. The 'squeezing in' of the graph reflects how what we're plugging into the function, i.e. \(x^2\), just increases at a faster rate than \(x\) itself does.

3.14159265359

  • Trendsetter
  • **
  • Posts: 183
  • Respect: +16
Re: Graphing question
« Reply #4 on: October 21, 2018, 11:20:58 pm »
0
Hey,

thanks for your response

I'm still not too sure because if I apply your logic to this question, the answers don't match up (i've linked it for your convenience)
I just had a look at Ruis answer and whoops I'm sorry I was mistaken for the second part. I apologise.

also Rui i don't get why you mean here

For \( y=f(x^2) \), you square the \(x\)-coordinate before applying the function to it. Since the square of any number is negative, effectively speaking anything that was to the left of the \(y\)-axis on the original graph gets discarded, because for something like \(f(x^2)\) you'll never end up plugging a number like -1 into \(f\).

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Graphing question
« Reply #5 on: October 21, 2018, 11:28:54 pm »
+1
I just had a look at Ruis answer and whoops I'm sorry I was mistaken for the second part. I apologise.

also Rui i don't get why you mean here

Basically the stuff towards the left of the \(y\)-axis reflects what happens, when we plug a negative value of \(x\) into our function \(f(x)\).

But when we start dealing with \( f(x^2)\), we're never gonna be 'taking f of a negative value' anymore. This is because the square of a real number can never be a negative number. So because we've arrived at something impossible, what was originally to the left of the \(y\)-axis on \(y=f(x)\), becomes completely useless when we're graphing \(y=f(x^2)\)

mikamika

  • Adventurer
  • *
  • Posts: 6
  • Respect: 0
Re: Graphing question
« Reply #6 on: October 21, 2018, 11:31:00 pm »
0
Massive thank you to the both of you!!!

Now that the 4u exam is only 3 days away, what would you suggest the plan of attack be? I've done HSC papers from 2017-2008, so are there any papers pre 2008 which have hard questions? I personally think i'm going to re-do questions I got wrong from the papers I've done

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Graphing question
« Reply #7 on: October 21, 2018, 11:37:43 pm »
0
Massive thank you to the both of you!!!

Now that the 4u exam is only 3 days away, what would you suggest the plan of attack be? I've done HSC papers from 2017-2008, so are there any papers pre 2008 which have hard questions? I personally think i'm going to re-do questions I got wrong from the papers I've done
As a rule of thumb, the further back you go the more harder questions you'll see. 2001-2004 all had some pretty solid questions at the very end.

Redoing questions that you've gotten wrong is also a good strategy for before the day of the exams.

It's up to you how you want to balance between these two strategies you've mentioned. For me, I was just too lazy to re-do questions I had already done, although the strategy is still effective

3.14159265359

  • Trendsetter
  • **
  • Posts: 183
  • Respect: +16
Re: Graphing question
« Reply #8 on: October 21, 2018, 11:39:16 pm »
0
Basically the stuff towards the left of the \(y\)-axis reflects what happens, when we plug a negative value of \(x\) into our function \(f(x)\).

But when we start dealing with \( f(x^2)\), we're never gonna be 'taking f of a negative value' anymore. This is because the square of a real number can never be a negative number. So because we've arrived at something impossible, what was originally to the left of the \(y\)-axis on \(y=f(x)\), becomes completely useless when we're graphing \(y=f(x^2)\)

ooh okay I get you!!

Massive thank you to the both of you!!!

Now that the 4u exam is only 3 days away, what would you suggest the plan of attack be? I've done HSC papers from 2017-2008, so are there any papers pre 2008 which have hard questions? I personally think i'm going to re-do questions I got wrong from the papers I've done

no problem!! I reckon you go over your mistakes from the papers you have done and attempt hard questions from recent trial papers from either companies or schools, what ever you can find

mikamika

  • Adventurer
  • *
  • Posts: 6
  • Respect: 0
Re: Graphing question
« Reply #9 on: October 21, 2018, 11:53:19 pm »
0
Thanks! I'll defs be looking into harder trial papers/2001-04 HSC q's

Also, for part iii of this graphs question (2008 HSC), how did they get the part for g(2-x) for x<1 ??
Like what I did was flip the graph and move it 2 spaces to the right for the domain, but I don't end up getting the answer...

Thanks again!

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Graphing question
« Reply #10 on: October 22, 2018, 12:03:42 am »
0
Thanks! I'll defs be looking into harder trial papers/2001-04 HSC q's

Also, for part iii of this graphs question (2008 HSC), how did they get the part for g(2-x) for x<1 ??
Like what I did was flip the graph and move it 2 spaces to the right for the domain, but I don't end up getting the answer...

Thanks again!
Those transformations look appropriate. Can you post up your answer somehow?

3.14159265359

  • Trendsetter
  • **
  • Posts: 183
  • Respect: +16
Re: Graphing question
« Reply #11 on: October 22, 2018, 12:07:26 am »
0
Thanks! I'll defs be looking into harder trial papers/2001-04 HSC q's

Also, for part iii of this graphs question (2008 HSC), how did they get the part for g(2-x) for x<1 ??
Like what I did was flip the graph and move it 2 spaces to the right for the domain, but I don't end up getting the answer...

Thanks again!

what you said sounds right. this is what I would do, lmk if it doesn't make sense or (if embarrassingly, its wrong)

mikamika

  • Adventurer
  • *
  • Posts: 6
  • Respect: 0
Re: Graphing question
« Reply #12 on: October 22, 2018, 12:13:18 am »
0
what you said sounds right. this is what I would do, lmk if it doesn't make sense or (if embarrassingly, its wrong)

this is right!
haha i transformed it with the wrong axis haha my bad