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October 09, 2025, 04:06:40 am

Author Topic: VCE Methods Question Thread!  (Read 5723594 times)  Share 

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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3450 on: January 02, 2014, 06:39:17 pm »
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I think you mean synthetic division, which is just a faster method of doing things, same result though :)

Yeah that's it. Would it be worth it learning that method?

psyxwar

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Re: VCE Methods Question Thread!
« Reply #3451 on: January 02, 2014, 06:49:18 pm »
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Yeah that's it. Would it be worth it learning that method?
It'll probably take you like a minute to learn, so why not?
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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3452 on: January 02, 2014, 07:07:39 pm »
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I'm not getting anywhere with these.
(5x+6)/4 = (100-4x)/(3) -6


6/(x) + 4 = 15/(x) - 3.
With this i keep getting 7/9 instead of 9/7 so I must be multiplying the wrong thing.
Can someone note the steps to see where I went wrong?

(sorry that I'm bad at writing equations out) :P

clueless123

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Re: VCE Methods Question Thread!
« Reply #3453 on: January 02, 2014, 07:24:19 pm »
+3
a)
Spoiler









etc

b)
Spoiler











« Last Edit: January 02, 2014, 07:44:27 pm by clueless123 »
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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3454 on: January 02, 2014, 08:05:05 pm »
+1

alchemy

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Re: VCE Methods Question Thread!
« Reply #3455 on: January 03, 2014, 12:03:53 pm »
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How important is learning matrices? Is it necessary to know it for other topics?

It's usually just briefed over in most schools. It's that thing that you learn when you only come across it, which I like :)

Nato

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Re: VCE Methods Question Thread!
« Reply #3456 on: January 03, 2014, 02:05:35 pm »
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hey guys, having some trouble with dilations.

one question was state the transformation which maps the graphs of   to

for:  to
now i did


and therefore it has been dilated by a factor of 5 from the x-axis. Well, according to the book, my answer is correct. But could someone please provide me with an intuitive and/or more efficient way of solving this.

i also need some assistance with to

thank-you
Class of 2014.

soNasty

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Re: VCE Methods Question Thread!
« Reply #3457 on: January 03, 2014, 02:31:12 pm »
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Find the third solution to the equation
if it is known that -2 and 5 satisfy the equation

mackintosh

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Re: VCE Methods Question Thread!
« Reply #3458 on: January 03, 2014, 03:01:37 pm »
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Find the third solution to the equation
if it is known that -2 and 5 satisfy the equation
Are you sure there isn't an x cubed somewhere ? As it is written there should only be  two solutions

nhmn0301

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Re: VCE Methods Question Thread!
« Reply #3459 on: January 03, 2014, 03:36:09 pm »
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hey guys, having some trouble with dilations.

one question was state the transformation which maps the graphs of   to

for:  to
now i did


and therefore it has been dilated by a factor of 5 from the x-axis. Well, according to the book, my answer is correct. But could someone please provide me with an intuitive and/or more efficient way of solving this.

i also need some assistance with to

thank-you
Hi,
af(x)    Dilation factor of a, parallel to y-axis (from x-axis)
f(ax)    Dilation factor of 1/a, parallel to x-axis(from the y-axis)
f(1/x) Dilation factor of a, parallel to x-axis (from y axis)
so for your question, just apply the same methods above: to will be a dilation of factor (5)^1/2 from the x-axis.
Hope this helps.
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soNasty

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Re: VCE Methods Question Thread!
« Reply #3460 on: January 03, 2014, 05:00:12 pm »
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Are you sure there isn't an x cubed somewhere ? As it is written there should only be  two solutions

Sorry!! It's

brightsky

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Re: VCE Methods Question Thread!
« Reply #3461 on: January 03, 2014, 05:14:40 pm »
+1
Let P(x) = 3x^3 + ax^2 - 33x - b.
P(-2) = 4a - b + 42 = 0.........(1)
P(5) = 25a - b + 210 = 0.......(2)

Solve (1) and (2) simultaneously:
a = -8 and b = 10

So, P(x) = 3x^3 - 8 x^2 - 33x - 10 = (x+2)(x-5)(3x+1)

Therefore, the other solution to the equation P(x) = 0 is x = -1/3.
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3462 on: January 03, 2014, 05:48:52 pm »
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Just make sure you learn them in the right order. Quadratics first, cubics second, quartics third, ...
Also make sure you know what a polynomial is and isn't before you start off. Makes life easier later on.

Yea but if i know a poloynimal is how will that help me?  Isn't a polynomial any number/expression that does not have a negative power?  even 3 is a polynomial because u can write it as 3x^0 so the degree is 0.
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brightsky

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Re: VCE Methods Question Thread!
« Reply #3463 on: January 03, 2014, 05:59:12 pm »
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Pretty much!

A polynomial is essentially any function which can be written in the following form:

P(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_2 x^2 + a_1 x + a_0

where a_n, a_(n-1), ..., a_2, a_1, a_0 are all real number constants and n is a non-negative integer.

Bear in mind that n cannot be infinity. Technically, a function like sin(x) can be expressed as a power series, but it nevertheless is not a polynomial since it does not have a 'highest power' (i.e. degree). Also bear in mind that the degree of 0 (which is a polynomial) is undefined, not 0.
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3464 on: January 03, 2014, 08:39:09 pm »
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Pretty much!

A polynomial is essentially any function which can be written in the following form:

P(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_2 x^2 + a_1 x + a_0

where a_n, a_(n-1), ..., a_2, a_1, a_0 are all real number constants and n is a non-negative integer.

Bear in mind that n cannot be infinity. Technically, a function like sin(x) can be expressed as a power series, but it nevertheless is not a polynomial since it does not have a 'highest power' (i.e. degree). Also bear in mind that the degree of 0 (which is a polynomial) is undefined, not 0.

The p(x)=…….  What does that actually mean am i suppose to memorise that?
'My belief is stronger than your doubt'