HSC Stuff > HSC Mathematics Extension 2

Graphing question

(1/3) > >>

mikamika:
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:

--- Quote from: mikamika on October 21, 2018, 10:32:37 pm ---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

--- End quote ---

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:
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:
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:

--- Quote from: mikamika on October 21, 2018, 10:48:40 pm ---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)

--- End quote ---
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


--- Quote from: RuiAce on October 21, 2018, 11:12:37 pm ---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 \(f\).

--- End quote ---

Navigation

[0] Message Index

[#] Next page