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Author Topic: VCE Methods Question Thread!  (Read 4802916 times)  Share 

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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17685 on: February 18, 2019, 08:45:07 pm »
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Anyone know how to factorise \(12x^2+x-6=0\) using a quick method other than trial and error? I really need to know how to complete these quicker.

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17686 on: February 18, 2019, 09:28:30 pm »
+1
Anyone know how to factorise \(12x^2+x-6=0\) using a quick method other than trial and error? I really need to know how to complete these quicker.

I think the only (viable) methods that don't use trial and error are completing the square and the quadratic formula, which can be very tedious at times.

There are some techniques you can employ however to make guessing a lot easier, especially when the coefficient of the \(x^2\) term is not equal to 1.

Let's use your example. First, multiply the coefficient of the \(x^2\) term (\(a\)) with the constant term (\(c\)). In this case, \(-72\). Therefore, we would like to find two numbers (let's call them \(m\) and \(n\)) so that  \(mn=-72\)  and  \(m+n=1\),  the coefficient of the \(x\) term (\(b\)). We will use these numbers to split up the \(x\) term.

In your case, we need the numbers \(9\) and \(-8\). So,\begin{align*}12x^2+x-6&=12x^2-8x+9x-6 \tag{split x term}\\
&=4x(3x-2)-3(3x-2) \tag{factorise terms in pairs}\\
&=(4x-3)(3x-2)\tag{factorise further}\end{align*}
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nxthxnsxx

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Re: VCE Methods Question Thread!
« Reply #17687 on: February 18, 2019, 11:02:29 pm »
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Hihihihi!!

I just started Maths Methods 1/2 this year and right from the "get-go", i've been pretty determined to make sure I do pretty well in Methods since it's a prerequisite for my course (I plan to get into Commerce at Monash University).

However, after a mere 3 weeks, I've quickly noticed how I'm struggling and in a constant state of confusion when I'm trying to answer quesitons and hence, I'm quickly starting to lose motivation and becoming increasingly stressed.

It's not to say i'm a bad maths student; I think i'm mediocre. I did end up finishing the top of Advance Maths II (2nd class to the top class) class and scoring the highest exam mark in my class last year, but I'm struggling to grasp the concepts and apply it to all sorts of situations.

Is it normal to feel slightly overwhelmed in the opening weeks of Maths Methods and what is some advice you can empart unto me to ultimately succeed in Methods?

Really appreciate any comments that you may have!!

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17688 on: February 18, 2019, 11:49:23 pm »
+1
Hihihihi!!

I just started Maths Methods 1/2 this year and right from the "get-go", i've been pretty determined to make sure I do pretty well in Methods since it's a prerequisite for my course (I plan to get into Commerce at Monash University).

However, after a mere 3 weeks, I've quickly noticed how I'm struggling and in a constant state of confusion when I'm trying to answer quesitons and hence, I'm quickly starting to lose motivation and becoming increasingly stressed.

It's not to say i'm a bad maths student; I think i'm mediocre. I did end up finishing the top of Advance Maths II (2nd class to the top class) class and scoring the highest exam mark in my class last year, but I'm struggling to grasp the concepts and apply it to all sorts of situations.

Is it normal to feel slightly overwhelmed in the opening weeks of Maths Methods and what is some advice you can empart unto me to ultimately succeed in Methods?

Really appreciate any comments that you may have!!

Hey there! Next time, this may be better as it's own topic (won't be surprised if the mods move it), but that's okay :)

To answer your overall question:  yes, it is completely normal to feel overwhelmed!

VCE Mathematics is quite a jump from year 10 maths, and you are not expected to understand concepts upon first presentation - heck - even the second or third time. In fact, I would be somewhat suspicious of anyone if they didn't feel the slightest bit overwhelmed in the beginning. VCE Maths is the first time in your maths education where you are not fully protected or hidden from concepts that could ultimately cause questions or problems. It'll also be the first time where you start developing mathematical tools with some degree of rigor. Therefore, it'll take some adjusting to become accustomed to VCE life.

I can understand how one can lose motivation to learn when they are confused, and I can understand that it can be stressful. But, you need to channel that energy and put it into improving. The first think you should do as soon as possible is identify all your points of confusion. Then, as I say to my students, "attack me and your teacher with questions". Use your teachers. If you are struggling, ask for help.

Also, you have an entire community of people here who are more than willing to help students like you ;)

I'll leave you with a few quotes. Make of them what you will.

> "Ask why. Ask how. Ask whether it makes sense. Ask what if. Ask what if not."

> "Mathematics reveals its secrets only to those who approach it with pure love, for its own beauty."

> "The only way to learn mathematics is to do mathematics."

> "Mathematicians aren't people who find maths easy. They're people who enjoy how hard it is."
« Last Edit: February 18, 2019, 11:51:53 pm by AlphaZero »
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EllingtonFeint

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Re: VCE Methods Question Thread!
« Reply #17689 on: February 19, 2019, 11:37:41 am »
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Hi,
I'm confused about how to find the answer to this question... I think I need a super quick refresh on how to draw a parabola and figure out turning points and stuff from an equation...  :-[

The problem in question is
Let g: [b, infinity) --> R, where g(x) = x^2 + 4x. If b is the smallest real number such that g has an inverse function find b and g^-1 (x).

I literally just put that into my CAS to find the smallest real number that b could be. (so, b is -4) but I don't remember how to do this irl without a cas....

Also, I'm not quite sure how to complete the rest of the question.
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17690 on: February 19, 2019, 12:06:39 pm »
+1
Hi,
I'm confused about how to find the answer to this question... I think I need a super quick refresh on how to draw a parabola and figure out turning points and stuff from an equation...  :-[

The problem in question is
Let g: [b, infinity) --> R, where g(x) = x^2 + 4x. If b is the smallest real number such that g has an inverse function find b and g^-1 (x).

I literally just put that into my CAS to find the smallest real number that b could be. (so, b is -4) but I don't remember how to do this irl without a cas....

Also, I'm not quite sure how to complete the rest of the question.

Actually, the smallest value of \(b\) is \(-2\), but let's find out why that is.

Recall that, for  \(g^{-1}\),  the inverse function of \(g\),  to exist,  \(g\) must be a one-to-one function.

If we think about the general shape of a positive quadratic, we see that in this case, the smallest possible value of \(b\) will actually be the \(x\)-coordinate of the turning point. (See the diagram below)



The local minimum of \(g\) occurs at \(x=\dfrac{-4}{2\times 1}=-2\), so the minimal value of \(b\) is \(-2\).\[g:[-2,\infty)\to\mathbb{R},\ g(x)=x^2+4x.\] Now, let's answer the second part of the question. For convenience, let \(y=g^{-1}(x)\). \begin{align*}x&=y^2+4x\\
\implies x&=(y+2)^2-4 \tag{complete the square}\\
\implies x+4&=(y+2)^2\end{align*} Note that, when we square root both sides, we have to take the positive root only, but why? Well, recall that  \(\text{range}(g^{-1})=\text{domain}(g)=[-2,\infty)\),  so  \(g^{-1}(x)+2\geq0\). Thus, \[g^{-1}(x)=\sqrt{x+4}-2.\] Except, we haven't fully answered the question. Note that we are asked to find \(g^{-1}\), not just its rule, so we need to supply a domain.

From the graph, it is easy to see that  \(\text{range}(g)=\text{domain}(g^{-1})=[-4,\infty)\). Thus, \[\boxed{g:[-4,\infty)\to\mathbb{R},\ g^{-1}(x)=\sqrt{x+4}-2.}\]
« Last Edit: February 19, 2019, 12:09:22 pm by AlphaZero »
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EllingtonFeint

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Re: VCE Methods Question Thread!
« Reply #17691 on: February 19, 2019, 07:48:55 pm »
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Actually, the smallest value of \(b\) is \(-2\), but let's find out why that is.

Recall that, for  \(g^{-1}\),  the inverse function of \(g\),  to exist,  \(g\) must be a one-to-one function.

If we think about the general shape of a positive quadratic, we see that in this case, the smallest possible value of \(b\) will actually be the \(x\)-coordinate of the turning point. (See the diagram below)

(Image removed from quote.)

The local minimum of \(g\) occurs at \(x=\dfrac{-4}{2\times 1}=-2\), so the minimal value of \(b\) is \(-2\).\[g:[-2,\infty)\to\mathbb{R},\ g(x)=x^2+4x.\] Now, let's answer the second part of the question. For convenience, let \(y=g^{-1}(x)\). \begin{align*}x&=y^2+4x\\
\implies x&=(y+2)^2-4 \tag{complete the square}\\
\implies x+4&=(y+2)^2\end{align*} Note that, when we square root both sides, we have to take the positive root only, but why? Well, recall that  \(\text{range}(g^{-1})=\text{domain}(g)=[-2,\infty)\),  so  \(g^{-1}(x)+2\geq0\). Thus, \[g^{-1}(x)=\sqrt{x+4}-2.\] Except, we haven't fully answered the question. Note that we are asked to find \(g^{-1}\), not just its rule, so we need to supply a domain.

From the graph, it is easy to see that  \(\text{range}(g)=\text{domain}(g^{-1})=[-4,\infty)\). Thus, \[\boxed{g:[-4,\infty)\to\mathbb{R},\ g^{-1}(x)=\sqrt{x+4}-2.}\]

I’m not too sure how to quote properly 😬 but I wanted to say THANK YOU!
Cos that does make a bunch more sense now!
I’ll revise completing the square and graphing quadratics :)


Actually, maybe if you get a moment could you please put in a more detailed explanation of how you completed the square in this problem please? 🙏🏼
« Last Edit: February 19, 2019, 08:33:10 pm by EllingtonFeint »
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KiNSKi01

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Re: VCE Methods Question Thread!
« Reply #17692 on: February 19, 2019, 08:57:12 pm »
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Yoo can i please have some help? Heres the question:

A newspaper determined that an approximate 95% confidence interval for the proportion of people in Australia who regularly read the newspaper online was (0.5,0.7) What is the margin or error?

Textbook says the answer is 0.10 but I got 0.098 (don't think its a rounding issue because nowhere does it state to certain number of DP)

Not sure why my answer is wrong  :-\
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17693 on: February 19, 2019, 09:47:59 pm »
+3
Yoo can i please have some help? Heres the question:

A newspaper determined that an approximate 95% confidence interval for the proportion of people in Australia who regularly read the newspaper online was (0.5,0.7) What is the margin or error?

Textbook says the answer is 0.10 but I got 0.098 (don't think its a rounding issue because nowhere does it state to certain number of DP)

Not sure why my answer is wrong  :-\

The margin of error of a confidence interval is defined as half the distance between the endpoints, or equivalently, the distance from the centre to each endpoint. Essentially, \[M=z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]In this case,  \(M=\dfrac{0.7-0.5}{2}=0.1\).
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KiNSKi01

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Re: VCE Methods Question Thread!
« Reply #17694 on: February 19, 2019, 10:45:32 pm »
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I used the first equation u provided and had p as 0.6 and n as 96 (from memory) and thats how I got 0.098. Have I stuffed up here/ why do you get a different answer?
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17695 on: February 19, 2019, 11:14:10 pm »
+1
I used the first equation u provided and had p as 0.6 and n as 96 (from memory) and thats how I got 0.098. Have I stuffed up here/ why do you get a different answer?

You would've had to use that equation or equivalent to obtain  \(\hat{p}=0.6\)  and  \(n=96\)  in the first place though.

You haven't necessarily 'stuffed up'. You essentially used the margin of error and that equation (or equivalent) to find \(n\), only to plug \(n\) and \(\hat{p}\) back into the same equation to find the margin of error. What you did is 'mathematically correct', but completely pointless. (Although I believe \(n\approx 92\), not 96).

Also, for a 95% confidence interval,  \(z\approx 1.96\),  and is not exact. If you type  \(\texttt{invNorm(0.975,0,1)}\)  into your CAS, you should find that in fact it gives you  \(1.9599639859915...\). This rounding error plus your error in calculating \(n\) would've led to your answer.
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KiNSKi01

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Re: VCE Methods Question Thread!
« Reply #17696 on: February 20, 2019, 06:25:30 pm »
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Ok thankyou. Realised I did waste my time but just wanted to check I understood the relationship between margin of errors and confidence intervals  ;D
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Re: VCE Methods Question Thread!
« Reply #17697 on: February 21, 2019, 09:08:25 pm »
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Hi guys, just have a question about transformations that I'd like some help with   :)

If a domain of a function is (-2,2)

The function then undergoes a series of transformations:

-   A translation of one unit in the negative direction of y-axis
-   A translation of one unit in the negative direction of the x-axis
-   A dilation by factor 1/3 from the x axis

What is the new domain? Does it even change?

Thanks!
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Jimmmy

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Re: VCE Methods Question Thread!
« Reply #17698 on: February 21, 2019, 09:26:08 pm »
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Hi guys, just have a question about transformations that I'd like some help with   :)

If a domain of a function is (-2,2)

The function then undergoes a series of transformations:

-   A translation of one unit in the negative direction of y-axis
-   A translation of one unit in the negative direction of the x-axis
-   A dilation by factor 1/3 from the x axis

What is the new domain? Does it even change?

Thanks!
Do you know what the function is? Parabola, Quartic, Inverse?  I don't think it will change the domain unless explicitly stated, but it may change the range of y-values you get in that domain.
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jinaede1342

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Re: VCE Methods Question Thread!
« Reply #17699 on: February 21, 2019, 09:29:10 pm »
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Do you know what the function is? Parabola, Quartic, Inverse?  I don't think it will change the domain unless explicitly stated, but it may change the range of y-values you get in that domain.

It's a parabola...that's what I thought too, I guess I'll just state the domain as being (-2,2)

thanks
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