ATAR Notes: Forum
HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: mikamika on October 21, 2018, 10:32:37 pm
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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
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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 :)
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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)
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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.
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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 \(f\).
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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 \(y=f(x^2)\)
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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
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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
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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
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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!
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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?
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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)
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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