VCE Methods Question Thread!

[cosine]

Im still struggling with working out x-intercepts with graphs like y=2cos(x/3)

Can someone show me step by step on how to do this, including the change of domain and why you do it also. Thanks

[Floatzel98]

Im still struggling with working out x-intercepts with graphs like y=2cos(x/3)

Can someone show me step by step on how to do this, including the change of domain and why you do it also. Thanks

Does that have a restricted domain or do you just want the general solutions for the intercepts?

[cosine]

Does that have a restricted domain or do you just want the general solutions for the intercepts?

Just for simplicity, let's restrict it to [0, 2pi]

Thank you

[Floatzel98]

Just for simplicity, let's restrict it to [0, 2pi]

Thank you

Firstly:




Then we change the domain. I usually sub the domain into whatever is inside the brackets to do it.

If we let , when we sub in 0 we get 0 and when we sub in we get

So we have to solve in the domain

We can already see we will only have the 1 postive cos solution which will be which i can assume you understand?

Then we can just go on like normal

, Therefore



Falls within 2pi so we all good. That should be all to it i think

[cosine]

When we graph a trig function and the question states to sketch one full cycle, when stating the domain, would we say all Real numbers or also restrict the domain to the one cycle period? Cheers

[keltingmeith]

When we graph a trig function and the question states to sketch one full cycle, when stating the domain, would we say all Real numbers or also restrict the domain to the one cycle period? Cheers

Depends on the very specific wording. If they say the function has relationship f(x)=sin(x), to sketch it for one period, and then write the maximal domain of f(x), you'd put R. If they say to sketch it for one period and state this restricted domain, it's [0, 2pi]. If they just say domain, check to see if they ask for the domain of your sketch, or the domain of the function.

We hit one of the first problems in VCE maths - wording is super critical, and one change in word can change your entire answer. Pay attention to the wording.

[Sundal]

Hi, I'm doing polynomials, remainder/factor theorem at the moment and I do not understand what this means:

If (x − a) is a factor of P(x) and a is an integer, then a must be a factor of the term independent of x. For example, if (x − 2) is a factor of P(x), then the term independent of x must be divisible by 2. Therefore, (x − 2) could be a factor of x3 − 2x2 − x + 2, but (x + 3) could not be a factor.

[wunderkind52]

Hi, I'm doing polynomials, remainder/factor theorem at the moment and I do not understand what this means:

If (x − a) is a factor of P(x) and a is an integer, then a must be a factor of the term independent of x. For example, if (x − 2) is a factor of P(x), then the term independent of x must be divisible by 2. Therefore, (x − 2) could be a factor of x3 − 2x2 − x + 2, but (x + 3) could not be a factor.

I believe it means the constant. So consider x^2 -x-6, if you want to factorise it, the you have to find numbers that 6 is divisible by. The factors are (x-3)(x+2). Notice that 6 is divisible by both 3 and 2. Because when you expand it, it becomes (x^2 -3x+2x -3*2).

[Sundal]

I believe it means the constant. So consider x^2 -x-6, if you want to factorise it, the you have to find numbers that 6 is divisible by. The factors are (x-3)(x+2). Notice that 6 is divisible by both 3 and 2. Because when you expand it, it becomes (x^2 -3x+2x -3*2).

Ahh, I get it now.
Thanks wunderkind52!

[Sundal]

How would I go about doing this question?

Cheers 

[wunderkind52]

How would I go about doing this question?

Cheers 

Split x^2-1 using DOPS (x-1)(x+1).
Then use factor theorem on both factors to find A and B.

[silverpixeli]

How would I go about doing this question?

Cheers 

If (x^2-1) is a factor then the linear factors of (x^2-1) are also factors of the polynomial. (x^2-1) factorises to (x+1)(x-1) using difference of two squares, meaning we've just been effectively told that (x-1) and (x+1) are both factors of P(x).
Using this info, we can set up two equations using the factor theorem (which says "P(n)=0 means (x-n) is a factor") and solve these two equations simultaneously for a and b.

[@#035;3]

Hey guys, quick questions...
1. x^2+y^2-6x=0 
How would I factorise this by completing the square? 

2. A circle has y-intercepts at (0,2) and (0,-6). If the centre lies on the line x=2, find it's equation.

Any help would be greatly appreciated.

[Callum@1373]

Hey guys, quick questions...
1. x^2+y^2-6x=0 
How would I factorise this by completing the square? 

2. A circle has y-intercepts at (0,2) and (0,-6). If the centre lies on the line x=2, find it's equation.

Any help would be greatly appreciated.

For the first one, you will have to complete the square on the x

so it's x^2 -6x + 9 - 9 + y^2 = 0

(x-3)^2 + y^2 = 9

centre at (3,0), radius = 3

Second question
So it's y intercepts are 8 apart
therefore subtract half of 8 = 4 from the (0,2) coordinate and you get (0,-2)

Therefore it is of the form x^2 + (y-2)^2 = r^2

the radius was 4 so x^2 + (y-2)^2 = 16

if the circle lies on x=2 then it is shifted 2 units in positive x axis

therefore equation is (x-2)^2 + (y-2)^2 = 16
 

[Sundal]

Thanks wunderkind52 and silverpixeli for the help!