VCE Methods Question Thread!

[abd123]

Yeah, the examiners are hard. Everyone knows that VCAA Examiners can be that tight asses sometimes.

I don't what you mean by 'using (a-b)^3' perhaps you mean rewriting that question again and solving?
I think you should rewrite the questions.

Also, if its like a short answer thats worth 3 or 4 marks, of course you have to show the solutions to get full marks, I might be wrong though. For you to be sure, you should try to look atleast any methods reports ~ Mathematical Methods Examination 1 (short answers).

[Phy124]

Are examiners harsh when it comes to factorising?
As in, do you have to say "Using (a-b)^3 = ...." before you factorise something, or can you simply write the answer?
Tbh, I don't really know how you can show steps for factorising lolol.

If that means what I think it does... I doubt that you would have to as you don't have the presence of a calculator (where you would usually have to show working out in non-exam conditions). By the time exams come around you will have done enough practice exams to know how much working out is required for the amount of marks the question is out of (if you know what I mean )

Yeah, the examiners are hard. Everyone knows that VCAA Examiners can be that tight asses sometimes.

I don't what you mean by 'using (a-b)^3' perhaps you mean rewriting that question again and solving?
I think you should rewrite the questions.

Also, if its like a short answer thats worth 3 or 4 marks, of course you have to show the solutions to get full marks, I might be wrong though. For you to be sure, you should try to look atleast any methods reports ~ Mathematical Methods Examination 1 (short answers).

I think Hutchoo is referring to how you would expand by subbing in the values from the question for a and b i.e.

(a-b)³ = -b³ + 3ab² - 3a²b + a³

Therefore (x-2)³ = -(2)³ + 3(x)(2)² - 3(x)²(2) + x³

[Hutchoo]

Haha, thanks Phy124, that's exactly what I meant.

[Deceitful Wings]

is it possible for a cubic function to have one turning point and one point of inflexion? if so, give the equation and draw the graph of an example of such a cubic. if not, explain why.

[pi]

No, its not possible. I'm not sure on a reason, but to have one stationary inflection point (I'm assuming you mean a "stationary" inflection) and a local max/min turning point you need an equation which would mean that the graph would have to be a quartic (or higher, as some quintics will also have these properties).

[paulsterio]

Well, the simple explanation is that a derivative of a cubic is a quadratic. We know a quadratic can have 2 solutions, 1 solution (perfect square) and 0 solutions.

If the derivative has 2 solutions, cubic will have 2 turning points (min/max)

If the derivative has 1 solution, cubic will have 1 stationary point of inflection

If the derivative has 0 solutions, cubic will have no stationary points

So no, you can't have one with an SPOI as well as a TP

(I'm assuming since you asked it in Methods, you mean Stationary Point of Inflection, not just any Point of inflection - make sure you include "Stationary" next time, especially on SACs/Exams)

[Planck's constant]

Well, the simple explanation is that a derivative of a cubic is a quadratic. We know a quadratic can have 2 solutions, 1 solution (perfect square) and 0 solutions.

If the derivative has 2 solutions, cubic will have 2 turning points (min/max)

If the derivative has 1 solution, cubic will have 1 stationary point of inflection

If the derivative has 0 solutions, cubic will have no stationary points

So no, you can't have one with an SPOI as well as a TP

(I'm assuming since you asked it in Methods, you mean Stationary Point of Inflection, not just any Point of inflection - make sure you include "Stationary" next time, especially on SACs/Exams)

I like the thinking here, except for the unecessary assumption of SPOI + Turning. The problem said POI + Turning point

I would modify your explanation as follows:

1st Derivative 0 solutions : as per your explanation. There are no SP's, let alone TP's
1st Derivative 1 solution : as per you explanation. The same x-value which makes the 1st derivative 0, also makes the 2nd derivative 0, so we are dealing with a cubic which has precisely one SPOI and no TP's
1st Derivative has 2 solutions. We now want one of these solutions to make the 2nd derivative 0 and the other solution to make the 2nd derivative non-zero. This means that one of the solutions of the 1st derivative must be repeating. Which implies a cubic as a 1st derivative. A contradiction.

[Phy124]

I thought POI's were only in the specialist course (I didn't do spesh, not sure) and were calculated by double the double derivative?

[Planck's constant]

I thought POI's were only in the specialist course (I didn't do spesh, not sure) and were calculated by double the double derivative?

Thats true.
Come to think of it, the only mention of POI's in MM 3/4 was with respect to a cubic in Turning Point form
I withdraw my previous post

[Kanon]

Afternoon all, I realize it's not the holidays (well i'm still in holiday mode ), but I don't think i have enough questions to warrant a new thread.

Find the equation of the parabola

So for this one i've come to the conclusion it's y = (x+6)^2 -20, which i'm pretty sure is correct as I graphed it in Wolfram Alpha and it looks exactly the same.

However, for the following question

Determine the co-ordinates of the points of intersection A and B

In the question, we're given "[the parabola] is intersected twice by the line y = -x - 20"
My question is, how do we find the point of interception?  If I remember correctly it involves using simultaneous eqn?  I've forgotten how you'd do Simulataneous Eqn with an x to the second degree though. 

[pi]

Rearrange, solve for and then re-plug for values. I think you'll manage with those hints

[Phy124]

Afternoon all, I realize it's not the holidays (well i'm still in holiday mode ), but I don't think i have enough questions to warrant a new thread.

Find the equation of the parabola

So for this one i've come to the conclusion it's y = (x+6)^2 -20, which i'm pretty sure is correct as I graphed it in Wolfram Alpha and it looks exactly the same.

However, for the following question

Determine the co-ordinates of the points of intersection A and B

In the question, we're given "[the parabola] is intersected twice by the line y = -x - 20"
My question is, how do we find the point of interception?  If I remember correctly it involves using simultaneous eqn?  I've forgotten how you'd do Simulataneous Eqn with an x to the second degree though. 

At the point of intersection their y values will be the same so:

(x+6)²-20 = -x -20
x² + 12x + 36 - 20=-x - 20
x² + 13x + 36 = 0






If you need the coordinate put those back into either equation and your done

edit: nvr mind pi already helped 

Moderator action: removed real name, sorry for the inconvenience

[Bhootnike]

hi,
was wonderning how u would generally solve a question where you are asked to find the equation/gradient of a straight line, y = mx + c, , which is tangent to a circle of formula .... etc etc?, using calculus?

eg,  find m in y=mx -1, if this line makes tangent to circle (x-2)^2 + (y-4)^2 = 4
using calculus, !

[Phy124]

See uploaded image here, I'm sorry it would take too long in latex

Alternatively, you can skip lines (on the left) 2-6, and just sub mx - 1 for y, it'll get you to the same spot - quicker too.

I'm not sure of any other methods, because if you were to differentiate both equations and make them equal to each other, the m would be taken away as the derivative of mx - 1 = x. Of course, I'm most likely wrong, but yeah 
^ I'll use strikeout and not delete for other people's humour purposes *fp*

Anyway, I can't really see a question like this appearing on a methods Exam 

[nubs]

because if you were to differentiate both equations and make them equal to each other, the m would be taken away as the derivative of mx - 1 = x. Of course, I'm most likely wrong, but yeah 

haha
sure the derivative of mx=x?
dw I agreed with you for a while before I re-read it

I did it a different way, using a bit of calculus but I got a different answer to you and I'm pretty sure I'm wrong, because I would've done exactly what you did if it didn't say, 'using calculus' and your working out looks perfect to me