2026 HSC CHAT (4 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
 

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.
 

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