2026 HSC CHAT (6 Viewers)

Study to success

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Are there are topic stuff you want I can check
Can also try generating a paper from past HSC questions
Thanks. Maybe some stuff on linear, non linear and polynomials cause I have a topic test on that first week back and maybe some stuff for term 2 like calculus. Also we should all put some more stuff on the drive cause it is kinda empty
 

NotBamboo

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domain:
1. solve the quadratic and identify its concavity to find the x values where it is negative e.g. if it has x intercepts at x = -2 and 2, and is concave up the quadratic is negative between x = -2 and 2. since u cant square root a negative and cant divide by 0, that gives u the domain (x < -2, x > 2)

range:
2. find the range of the quadratic by finding the y coord of the vertex. e.g. vertex at (1, 3) and concave down = range: y ≤ 3
3. square root that (keeping in mind u can't square root a negative)-> 0 ≤ y ≤ sqrt3
4. find the reciprocal of that (a graph of y = 1/x could be useful - imagine the range of 1/x if you could only plug in values between 0 and sqrt3 (from step 3) -> y ≥ 1/sqrt3
silly she needed help with the ii)

also can u show me a faster way of doing it, refer to previous post
if this in my exam its gonna use a buncha time
 

cristal

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my working?
i even had to clean it up and add lil comments for u guys
ur welcome

i can explain any part of it if u dont understand
I don't understand anything 💀
I'll work through it first myself just to understand
But why differentiation btw?
 

NotBamboo

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I don't understand anything 💀
I'll work through it first myself just to understand
But why differentiation btw?
1745113151942.png
At one point the gradient is 0
When u differentiate it, the y value of it will be 0 because original function's gradient is 0, making the x easy to find.
This is called a critical point
Usually, u would use the second derivative to find the nature of the turning point (being a max)
But here since I already know a vague shape of the graph, I know that this will be the local maximum, so I only need to use first derivative
now that i have the x value by subbing in y=0, i can plug it into the original function to get the y value of the point as well yayyyyy!!!!
THE END.
 

cristal

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Omg the water looks rlly nice 🤩😍


@NotBamboo I still don't understand 💀 😥
Idk everything much abt derivatives yet and all the first and second stuff and I don't know what's a local maximum is (do you mean like the turning point)

Also it's only like chapter 3 right you wouldn't really be using that yet right
 

NotBamboo

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Omg the water looks rlly nice 🤩😍


@NotBamboo I still don't understand 💀 😥
Idk everything much abt derivatives yet and all the first and second stuff and I don't know what's a local maximum is (do you mean like the turning point)

Also it's only like chapter 3 right you wouldn't really be using that yet right
wait oopsies
second derivative is in 12 adv
intro to calculus should already be taught in 11 adv right?
like derivatives and stuff
critical point local max local min -> just think of them as turning points they basically are
 

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