VCE Methods Question Thread!

[Corey King]

My bad, I was thinking root may have meant something completely different.

You had me with roots for a bit, until here "Hopefully things make more sense with this clarification - all parabolas have two roots (either two distinct roots, two duplicate roots or two complex roots (but this is a whole different story))."

I know the x-intercept terminology

Finding the vertex makes sense. I got the x-intercepts but didnt know what to do with them, if anything

I forgot that inverse functions have to be one-to-one, and cannot be many-to-one. Since the inverse would turn the many-to-one into one-to-many.

We still haven't really learned why it can't be many to one. It doesn't seem too bad just having two inputs per output

Thank you for the effortful and detail response Jirachi.

[Bluebird]

Hey AN!
Can I have some help with this question? I've tried putting into my CAS:
f(0.75)=3.1
f(1.5)=0
f(0)=0
but I don't know if my values are even right because my calculator just says false.
Thank you in advance!

[1729]

Hey AN!
Can I have some help with this question? I've tried putting into my CAS:
f(0.75)=3.1
f(1.5)=0
f(0)=0
but I don't know if my values are even right because my calculator just says false.
Thank you in advance!

I don't understand what you're doing to solve this problem, but this is how I would solve the problem.

These are the points you have: (0,0) (1.5,0) and (0.75,0.6)

Note: You know the value of c is 0 since the parabola passes the origin.

y = a(x^2 - 1.5x)
0.6 = a(0.75^2 - 1.5(0.75))
0.6 = a(0.5625 - 1.125)
0.6 = -0.5625a
a = -16/15
now you can do...
-16/15(x^2 - 1.5x) and distribute and that is your funtion

Also I think your error was where you thought (0.75,3.1) was a cordinate. Cordinate of vertex is actually (0.75, 0.6) because 3.1 - 2.5

Hope this helps

[Corey King]

Hey guys
Do any of you know if this old TI calculator will be able to cover all the functions needed for the current study design?

[The Cat In The Hat]

Hey guys
Do any of you know if this old TI calculator will be able to cover all the functions needed for the current study design?

I believe not. That's what I used to have before I inherited my brother's but he'd had to get the newer one for Methods in any case. I'd at least strongly advise against it since you'd not have any guide for some things and I don't think you can even do all the functions required. I'd advise you to replace it with a newer one instead.

[Corey King]

Damn, thought that might be the case :/
How new does it need to be?

[The Cat In The Hat]

Damn, thought that might be the case :/
How new does it need to be?

I'm afraid I don't know. Probably ask your teacher. And also look secondhand if you want not quite such a horrid price.

[Bluebird]

I don't understand what you're doing to solve this problem, but this is how I would solve the problem.

These are the points you have: (0,0) (1.5,0) and (0.75,0.6)

Note: You know the value of c is 0 since the parabola passes the origin.

y = a(x^2 - 1.5x)
0.6 = a(0.75^2 - 1.5(0.75))
0.6 = a(0.5625 - 1.125)
0.6 = -0.5625a
a = -16/15
now you can do...
-16/15(x^2 - 1.5x) and distribute and that is your funtion

Also I think your error was where you thought (0.75,3.1) was a cordinate. Cordinate of vertex is actually (0.75, 0.6) because 3.1 - 2.5

Hope this helps

Thank you! I get it now

[a weaponized ikea chair]

where's everyone up to in class?

We're in chapter 4.

[pans]

where's everyone up to in class?

We're in chapter 4.

same. I know a lot of people r doing vectors tho

[Corey King]

We are finishing chapter 3 still.

[Ruchir]

Hello, I need help with this question
Answer for 6b is 430000, can anyone explain me how and why that is correct?

[parieeelol]

Just a really standard question about set notation/range -

For this picture, my answer for range was (0, ∞) or just R+
Could anyone explain why the book's answer uses the union symbol? Does this mean that the range is only either positive or zero?

[fun_jirachi]

The set union means that it includes both sets. Any two sets \(A, B\) defined by the relation \(A \cup B\) will include all items in set \(A\), set \(B\) and items in both sets. In our case, it means that the range of the function includes positive reals and zero (since no real number lies in both the positive reals and the set including only zero). It's effectively the same as the range \([0, \infty)\) but not the same as \(\mathbb{R}^+\) as the latter does not include zero (but the graph clearly does).

Hope this clears things up

[Ruchir]

Can anyone explain me Q7a
Why is the answer 0<k<1/4
And not just k<1/4