2026 HSC CHAT (5 Viewers)

Study to success

Well-Known Member
Joined
Sep 24, 2024
Messages
661
Location
my grave
Gender
Female
HSC
2026
View attachment 47381
HELPP I CANT DO THESEE AHSHFHFJ
I think there supposed to be like a semicircle/ hyperbola thing. Maybe try graphing it on Desmos so u have a rough idea on how it’s supposed to look. I think maybe finding the vertex and intercepts of the equation in the denominator can help u find the domain and range
 

cheesynooby

Well-Known Member
Joined
Feb 7, 2025
Messages
221
Location
punklorde
Gender
Male
HSC
2025
View attachment 47381
HELPP I CANT DO THESEE AHSHFHFJ
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
 
Last edited:

cheesynooby

Well-Known Member
Joined
Feb 7, 2025
Messages
221
Location
punklorde
Gender
Male
HSC
2025
So I’ll never be asked this in an adv exam right?
well it's adv content so theoretically u could be asked it
however it's very challenging and it's in the enrichment section, so something like this would only be asked if the examiners are feeling nasty :eek2: and would likely be split into many parts and be worth a lot of marks
 

NotBamboo

Well-Known Member
Joined
Apr 2, 2025
Messages
174
Location
Gender
Male
HSC
2026
1. if u solve the quadratic and identify its concavity u can 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)
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) -> y ≥ 1/sqrt3
bro i was gonna do it first 😭😭😭
cheater
 

Study to success

Well-Known Member
Joined
Sep 24, 2024
Messages
661
Location
my grave
Gender
Female
HSC
2026
well it's adv content so theoretically u could be asked it
however it's very challenging and it's in the enrichment section, so something like this would only be asked if the examiners are feeling nasty :eek2: and would likely be split into many parts and be worth a lot of marks
Ok. I’ll prob practice smth like this towards prelim or next year
 

Users Who Are Viewing This Thread (Users: 2, Guests: 3)

  • Top