VCE Methods Question Thread!

[AirLandBus]

To memory, neither is specifically stated in the study design. However, they have so many tougher things they can ask, as opposed to "factorise this". No, the only reason you learn about this stuff is to graph the functions by hand, I doubt VCAA will ever specifically ask about factorising, rather going straight for the "graph f(x)".

Oh cool. Im already very comfortable with factorising by long division so as long as its safe to do so, in which case it seems so, wont bother with the inspection method.

[lolaishappy]

In the exam or sacs where you have to find turning points (polynomials chapter) to sketch a graph without CAS. Will they ever ask to turn a equation to turning point form? I find it easier to find he derivativr, let it equal 0, solve the x-values, sub into original equation.

Thises Calculus, would any marks be deducted?

[brightsky]

Oh cool. Im already very comfortable with factorising by long division so as long as its safe to do so, in which case it seems so, wont bother with the inspection method.

Although you might be comfortable with polynomial long division,  it may still be rewarding for you to make an attempt at understanding the minutiae of what people have christened the inspection method. It can be quite dangerous to fixate on a particular way of doing things, especially when the method upon which you insist is less efficient than other available methods. The logic behind the inspection method is much the same as that behind polynomial long division, so investing some time into learning the inspection method might help to cast new light upon polynomial long division. Also, from a purely pragmatic point of view, it is always handy to have another method up your sleeve for the purposes of checking.

[keltingmeith]

In the exam or sacs where you have to find turning points (polynomials chapter) to sketch a graph without CAS. Will they ever ask to turn a equation to turning point form? I find it easier to find he derivativr, let it equal 0, solve the x-values, sub into original equation.

Thises Calculus, would any marks be deducted?

Whilst SAC standards are set by VCAA, the type of questions between each school is different, and the marking for them is also different, so we cannot answer any specific questions for SACs. For an exam, it is assumed you know the whole study design, and so they won't discount any points unless they specifically ask for a technique.

However, I'm assuming you're talking about quadratics here, and I feel there are several better methods to being able to draw them than using calculus. I'd definitely suggest you practice completing the square (even "completing the cube"), as it is a very useful tool. Even if you don't do that, factorising in order to sketch, and even use of the formula x=-b/2a to find the turning point instead. There's no need to reinvent the wheel, here.

[AirLandBus]

Hey Euler, quick question.

When sketching cubics by hand, say f(x)=-(x+2)^3 - 1

When i go to find the x intercept - let f(x) = 0 and i get two x intercepts, why is this? In which one is only applicable due to whether or not the graph has been reflected in any form. Is there any way to show which one is the correct x intercept in the case of the graph? Or am i missing something here.

[keltingmeith]

Hey Euler, quick question.

When sketching cubics by hand, say f(x)=-(x+2)^3 - 1

When i go to find the x intercept - let f(x) = 0 and i get two x intercepts, why is this? In which one is only applicable due to whether or not the graph has been reflected in any form. Is there any way to show which one is the correct x intercept in the case of the graph? Or am i missing something here.

You've made an error. I'm going to hazard a guess and say you added a +/- when you took the cube root?

[AirLandBus]

You've made an error. I'm going to hazard a guess and say you added a +/- when you took the cube root?

YEAH, bingo. Thanks man.

[keltingmeith]

Don't worry, it happens to the best of us.

[dankfrank420]

Convert x^3 - 9x^2 + 27x - 22 into A(x + B)^3 + C

Never heard of "completing the cube" before. I did what lolaishappy suggested above. I differentiated the polynomial, let it equal 0 to find the x-value stationary point and so on. Seeing as this will probably be asked in a SAC (and we haven't covered calculus yet), how would I find the answer without calculus? I'd probably lose full marks otherwise.

[pi]

(and we haven't covered calculus yet)

You didn't cover it in Methods 1+2?

[keltingmeith]

Convert x^3 - 9x^2 + 27x - 22 into A(x + B)^3 + C

Never heard of "completing the cube" before. I did what lolaishappy suggested below. I differentiated the polynomial, let it equal 0 to find the x-value stationary point and so on. Seeing as this will probably be asked in a SAC (and we haven't covered calculus yet), how would I find the answer without calculus? I'd probably lose full marks otherwise.

We know from the binomial theorem:



So, all we need to do is find that value of a, and adjust the constant term as necesarry. (which, believe it or not, is actually what you're doing when you complete the square)

From the above, we check the second term and divide by three, giving us 3. So, we'd assume a=3. Checking the third term, we get 27/3=9, and 3^2=9, so a=3 fits the bill. So now, we just need to make the constant term 3^3=27. So, this means we have:



Remember that because the second term had a minus, we had to make the constant have a minus for it to fit the binomial theorem.

[brightsky]

x^3 - 9x^2 + 27x - 22 = (x - 3)^3 + 27 -22 = (x-3)^3 + 5

Completing the cube is in general quite hard to do. You really need to be able to look at the cubic expression, think of a perfect cube that sort of resembles the expression, and then make some adjustments so that expressions coincide.

[lolaishappy]

Convert x^3 - 9x^2 + 27x - 22 into A(x + B)^3 + C

Never heard of "completing the cube" before. I did what lolaishappy suggested above. I differentiated the polynomial, let it equal 0 to find the x-value stationary point and so on. Seeing as this will probably be asked in a SAC (and we haven't covered calculus yet), how would I find the answer without calculus? I'd probably lose full marks otherwise

Like Euler said you can use :  x=-b/2a for the x-value, and then sub this into your equation for the y-value

[dankfrank420]

Thanks guys.

You didn't cover it in Methods 1+2?

I meant in the sense that we haven't learnt it in 3/4 methods yet. My first SAC will be on chs. 1-4, which doesn't cover calculus, so if I use a calculus technique to find an answer a different way to what I was taught from chs. 1-4 I might be penalised.

[keltingmeith]

Like Euler said you can use :  x=-b/2a for the x-value, and then sub this into your equation for the y-value

Important distinction:
For a cubic, this won't give you a turning point - this will only give you the point of inflection. Which for sketching a cubic than can be expressed in turning-point form, is all you need, but you should never make this assumption unless you can put it in turning point form.

Also note that the -b/2a refers to its derivative f'(x)=ax^2+bx+c, where f(x)=dx^3+ex^2+gx+h.