VCE Methods Question Thread!

[deStudent]

Thanks again Shadowxo,I owe you so much lol!

But wasn't what we found in (ii) A =... so doesn't this mean we want the domain of A?

[tasmia]

Need some general advice:
I've been using the Cambridge 3/4 book over the holidays with my tutor and we've already gone through the first area of studies. My school is using the jacaranda book (which I also have) and I've been told that we'll be doing the first area of studies throughout the whole term. Should I do the exercises from the book that the school is using or should I just ignore it all and do practice papers and move along to the next topic? Thanks

[undefined]

http://imgur.com/M4omKOF

Need some help with graphing this question. In the answers, it says the circle has a radius of 5, but isn't the square root of 169 13, therefore the radius be 13?

[Sine]

http://imgur.com/M4omKOF

Need some help with graphing this question. In the answers, it says the circle has a radius of 5, but isn't the square root of 169 13, therefore the radius be 13?

yeah radius is 13

Just back yourself or check on the CAS 

[Shadowxo]

Thanks again Shadowxo,I owe you so much lol!

But wasn't what we found in (ii) A =... so doesn't this mean we want the domain of A?

Yes but A won't exist unless x and y are both positive, so you have to find where A exists. If y and x are both positive, A definitely exists but if you find where A is positive, it may not exist as y will be negative, which is impossible. Does that help a bit?

Need some general advice:
I've been using the Cambridge 3/4 book over the holidays with my tutor and we've already gone through the first area of studies. My school is using the jacaranda book (which I also have) and I've been told that we'll be doing the first area of studies throughout the whole term. Should I do the exercises from the book that the school is using or should I just ignore it all and do practice papers and move along to the next topic? Thanks

Since you've already gone through it I'd recommend just doing a few questions from each chapter - this gives you a bit of revision while not going over it too much. Use it as well as practice papers to increase your speed, and especially do the more difficult worded ones.

http://imgur.com/M4omKOF

Need some help with graphing this question. In the answers, it says the circle has a radius of 5, but isn't the square root of 169 13, therefore the radius be 13?

Yes you're right. The radius should be 13 and the centre at (-1,4). If the book says the radius is 5, it's incorrect

Hope this helps all of you 

[undefined]

how do you find the domain and range of these types of q's?

[deStudent]

^^ plot the graph and observe. Would be all reals excluding the values of the asymptotes.

http://m.imgur.com/Q4024ky Ans has C, but it's A isn't it? They're implying that that both intercepts are +Ve?

[Sine]

how do you find the domain and range of these types of q's?

So the graph is a hyperbole
I assume you know the general idea of domains and ranges. (ask if you are unsure)
To find the domain pretty much find the numbers which cannot be input into the function.(this is usually due to fractions/logs) This function has x+2 in the denominator. We know the a denominator cannot =0. Thus x+2=/=0 and x=/=-2 So the domain cannot include -2 so domf: R\{-2}. (f(x)=y)

Range is what outputs are possible. Knowing it's a hyperbole we know from experience that this is R and not including one value. With experience you can read it off and automatically say ranf: R\{1}. Or consider what happens when x-->infinity. The denominator basically becomes infinity and thus -1/x+2 ---->0 and y--->1 so asymptote at y=1

[Shadowxo]

^^ plot the graph and observe. Would be all reals excluding the values of the asymptotes.

http://m.imgur.com/Q4024ky Ans has C, but it's A isn't it? They're implying that that both intercepts are +Ve?

This is a semi-common trick question.
y = 0 when one of the factors equals zero, so y equals zero when x=c so x-c = 0, so x-c is a factor.
Regardless of whether the intercepts are positive or negative, it will be A(x-a)(x-b)(x-c) where a, b, c are the x intercepts.
eg: c = -4. if x = -4 then x + c = -4 - 4 = -8.   x - c = - 4 + 4 = 0
Also, it has (b-x)² because (x-b)² = (b-x)² as a number squared is always positive.

Hope this helps a bit feel free to ask for clarification

[undefined]

can I get a step by step process on how to rearrange these? thanks

[Shadowxo]

can I get a step by step process on how to rearrange these? thanks

First you need to get rid of the denominator, then take out x and rearrange for x

Same for the second question, multiply all of the fractions by each of the denominators to get rid of them, then solve

Hope this helps

[codebreaker1_91]

How do I sketch this trigonometric graph?

[0,3pi/2] -> f(x)= -6sin(3x - 3pi/4)

Thanks

[Shadowxo]

How do I sketch this trigonometric graph?

[0,3pi/2] -> f(x)= -6sin(3x - 3pi/4)

Thanks

Trig graphs aren't my specialty, but I'll try and help you out
You know the shape of the graph, but I'd recommend finding the relevant points first as you need to know them.

You're given a domain so you need to find the endpoints. Sub in x=0 and x= 3pi/2 into f(x) to find the end points, these will also help you sketch the graph. Other things you need to know when sketching are the range - in this case [-6,6], the period, 2pi/n = 2pi/3, and x and y intercepts. You have the y intercept as that's an end point, but sub in y=0 to find the x intercept/s within the domain. From all this information you should be able to graph it, but sometimes you'll need to show a max/min point, so find where f(x) = 6 and -6 one time: solve (3x-3pi/4) = -π/2 for the former and π/2 for the latter. You should be able to sketch the graph from here with all the relevant points.

For just the general shape without solving for the points, we know the general shape is -sin(x), dilated 6 from the x axis, with a period 1/3 of the initial one and translated π/4 right.
Hope this helps a bit

[Ellise.M]

How do you find the implied domain of this function? I get that  x(2)+3 needs to be greater or equal to 0, but not sure how to get around squaring a negative number

[lzxnl]

How do you find the implied domain of this function? I get that  x(2)+3 needs to be greater or equal to 0, but not sure how to get around squaring a negative number

That is what you want. However, that's satisfied by all x, right? So the answer is simply all reals.