VCE Methods Question Thread!

[PolySquared]

Hey guys,

How do I graph something on the Ti-nspire cx Cas when points are given to you.

I have a question that says graph on the cas using the point (-3,5) (2,3)(1,-7)

[VanillaRice]

Hey guys,

How do I graph something on the Ti-nspire cx Cas when points are given to you.

I have a question that says graph on the cas using the point (-3,5) (2,3)(1,-7)

You'd first need to know what type of graph/function it is (exponential, quadratic, linear....). Is it stated in the question?

[DBA-144]

You'd first need to know what type of graph/function it is (exponential, quadratic, linear....). Is it stated in the question?

 

I think the function would be quadratic?

So all u need to do is sub each value into general form of quadratic, use that to make 3 simultaneous equations and the values u get for a b and c, u just put back into general form of quadratic.

Hope this helps

[Sine]

It seems to be quadratic but defintiely no one can be 100% sure because even though a function may look quadratic locally it could very well be something else.

[snowisawesome]

It seems to be quadratic but defintiely no one can be 100% sure because even though a function may look quadratic locally it could very well be something else.

Could we get a question in the methods exams (either exam 1 or 2) that gives us 3 points and asks us to graph them but doesn't specify what type of graph it is (ie quadratic, cubic, etc)?

[Lear]

Could we get a question in the methods exams (either exam 1 or 2) that gives us 3 points and asks us to graph them but doesn't specify what type of graph it is (ie quadratic, cubic, etc)?

Almost definitely no. There are many functions that can satisfy three points and this would lead to a range of possible answers. Therefore it is unlikely that vcaa would give us an open ended questions such as this without specifying type of graph.

[PolySquared]

You'd first need to know what type of graph/function it is (exponential, quadratic, linear....). Is it stated in the question?

It is a quadratic equation

[VanillaRice]

It is a quadratic equation

I'm not 100% sure if there's a way to immediately graph a quadratic equation given certain coordinates (although you can do a quadratic regression, and try to plot that somehow). Regardless, this wouldn't be a particularly useful skill to have in an exam/SAC situation, so I wouldn't worry about it too much. If you want to find a suitable equation to fit your points, you can solve simultaneous equations like DBA-144 has suggested.

So all u need to do is sub each value into general form of quadratic, use that to make 3 simultaneous equations and the values u get for a b and c, u just put back into general form of quadratic.

Hope this helps

[Trullzi]

Dunno, if im doing this right. But Hi everyone
I have a quick question regarding Composite Functions, it was on a Methods sheet i obtained from the VCE notes sections here. Here goes

f:R-->R, g:R-->R, g(x) = 2x^2 and f(g(x)) = 6x2-1, find f(x)
Idk if im just really dumb here for not noticing something, but let me know
I started with the idea that the range of g(x) would have to be a subset of domain of g(x) but, im unsure if thats even correct.

[RuiAce]

Dunno, if im doing this right. But Hi everyone
I have a quick question regarding Composite Functions, it was on a Methods sheet i obtained from the VCE notes sections here. Here goes

f:R-->R, g:R-->R, g(x) = 2x^2 and f(g(x)) = 6x2-1, find f(x)
Idk if im just really dumb here for not noticing something, but let me know
I started with the idea that the range of g(x) would have to be a subset of domain of g(x) but, im unsure if thats even correct.

\begin{align*}f(g(x))&=6x^2 - 1\\ &= 3 (2x^2) - 1\\ &= 3(g(x)) - 1\\ \therefore f(x) &= 3x-1\end{align*}

[Trullzi]

\begin{align*}f(g(x))&=6x^2 - 1\\ &= 3 (2x^2) - 1\\ &= 3(g(x)) - 1\\ \therefore f(x) &= 3x-1\end{align*}

Sorry could you explain why you factored 3 out? Is it due to g(x)= 2x^2?

[RuiAce]

Sorry could you explain why you factored 3 out? Is it due to g(x)= 2x^2?

Yeah. That was all the intuition that was necessary.

[Trullzi]

Yeah. That was all the intuition that was necessary.

Ah my bad, thanks for the help!

[snowisawesome]

am i correct in saying that if
-y = y' + 0.5
then y = -y' - 0.5

[RuiAce]

am i correct in saying that if
-y = y' + 0.5
then y = -y' - 0.5

Yeah.

Why wouldn't you be?