VCE Methods Question Thread!

[knightrider]

say if you were asked to put a cubic equation into the form how would you go about doing this?

[dankfrank420]

Knowing basic things like areas, volumes, basic angles (in relation to polygons, or directions on maps and so forth), and the concept of "similar shapes" are probably it.

From a brief look at the exams, you might also need to know surface areas of cones and boxes for calculus

[dankfrank420]

say if you were asked to put a cubic equation into the form how would you go about doing this?

Differentiate the polynomial, let it equal 0 and this gives you b. Sub the b value into the original polynomial, this gives you the c value. Now, you can write out the equation with a instead of a number.

I think you can solve from there.

[keltingmeith]

say if you were asked to put a cubic equation into the form how would you go about doing this?

Alternatively to the calculus method, you can adjust the constant term by inspection. Take, for example, y=3x^3-18x^2+36x-21.

1) Get rid of that coefficient in front of the x^3:
y=3(x^3-6x^2+12x-7)

2) Work the binomial theorem backwards to see if you can figure out what the b term might be. Looking at the second term, I'd assume it's -6/3=-2. (-2)^2=4=12/3, so that suits the third term. This means the last term should be (-2)^3=-8, so we fix that:
y=3(x^3-6x^2+12x-7-1+1)=3(x^3-3*2x^2+3*4x-8+1)

3) Finally, factorise:
y=3(x^3-6x^2+12x-8)+3
y=3(x-2)^3+3

[IndefatigableLover]

Knowing basic things like areas, volumes, basic angles (in relation to polygons, or directions on maps and so forth), and the concept of "similar shapes" are probably it.

From a brief look at the exams, you might also need to know surface areas of cones and boxes for calculus

@knightrider: More or less everything you'd need to know in terms of geometry for Methods will be given to you on your formula sheet (apart from the really basic ones).

[keltingmeith]

@knightrider: More or less everything you'd need to know in terms of geometry for Methods will be given to you on your formula sheet (apart from the really basic ones).

Also, those pesky similar triangles aren't on there - one of the top three things hated by nearly every methods student ever.

[Zues]

i know how to solve the second screenshot, but they didnt mention anything like the largest set of S to produce an inverse function. how they have said it can mislead me to to put its domain, rather then restricting it?

also with the graph of y = loge(x^2)  and (logex)^2, why is graphed the way it is? i.e. with the cusp at 0,0 (if thats what you call it)

[lzxnl]

Differentiate the polynomial, let it equal 0 and this gives you b. Sub the b value into the original polynomial, this gives you the c value. Now, you can write out the equation with a instead of a number.

I think you can solve from there.

Alternatively to the calculus method, you can adjust the constant term by inspection. Take, for example, y=3x^3-6x^2+12x-21.

1) Get rid of that coefficient in front of the x^3:
y=3(x^3-2x^2+4x-7)

2) Work the binomial theorem backwards to see if you can figure out what the b term might be. Looking at the second term, I'd assume it's -2. (-2)^2=4, so that suits the third term. This means the last term should be (-2)^3=-8, so we fix that:
y=3(x^3-2x^2+4x-7-1+1)=3(x^3-2x^2+4x-8+1)

3) Finally, factorise:
y=3(x^3-2x^2+4x-8)+3
y=3(x-2)^3+3

I'm happy with the calculus method tbh. That plus adjusting the leading coefficient by inspection should be the fastest method available

Zues, ln x^2 is undefined at x=0. So is (ln x)^2

Also, those pesky similar triangles aren't on there - one of the top three things hated by nearly every methods student ever.

Pffft. I hate those 'how many of the following are correct' questions. You have to then check EVERY single mc option

[keltingmeith]

... Actually, they did say that. That's pretty much EXACTLY what they said. Are we looking at the same screenshot?

Erm... Care to show us an example of those two graphs? Because neither of them should be defined for x=0.

[knightrider]

Differentiate the polynomial, let it equal 0 and this gives you b. Sub the b value into the original polynomial, this gives you the c value. Now, you can write out the equation with a instead of a number.

I think you can solve from there.

What happens if when you solve for b you get more than one value.

[Splash-Tackle-Flail]

Differentiate the polynomial, let it equal 0 and this gives you b. Sub the b value into the original polynomial, this gives you the c value. Now, you can write out the equation with a instead of a number.

I think you can solve from there.

How does subbing the b value into the original polynomial give the c value? And where do you sub in b? Isn't the b in y=a(x-b)^3+c different to the b in ax^3+bx^2+c?

[keltingmeith]

What happens if when you solve for b you get more than one value.

Then your cubic cannot be put into turning point form. If you get more than one value, then you have more than one stationary point - a cubic that can be placed in turning point form should only have one stationary point.

How does subbing the b value into the original polynomial give the c value? And where do you sub in b? Isn't the b in y=a(x-b)^3+c different to the b in ax^3+bx^2+c?

Remember that the T.P cubic has only one stationary point. So, we can abuse this fact by differentiating, and finding that stationary point. Once we've found it, we let the b in y=a(x-b)^3+c be that stationary point, because x=b is where the stationary point should occur (based on translational information)

Pffft. I hate those 'how many of the following are correct' questions. You have to then check EVERY single mc option

Never heard that one, before. I've only heard complaints about drawing graphs, probability and similar triangles.

[knightrider]

Alternatively to the calculus method, you can adjust the constant term by inspection. Take, for example,

1) Get rid of that coefficient in front of the x^3:


2) Work the binomial theorem backwards to see if you can figure out what the b term might be. Looking at the second term, I'd assume it's -2. (-2)^2=4, so that suits the third term. This means the last term should be (-2)^3=-8, so we fix that:


3) Finally, factorise:

yep but when you put it in the form isnt b,c the point of inflection so in your case it would be 2,3 which isnt the point of inflection

[keltingmeith]

yep but when you put it in the form isnt b,c the point of inflection so in your case it would be 2,3 which isnt the point of inflection

Firstly, a distinction - in the form y=a(x-h)^3+k, (h, k) is the STATIONARY point of inflection. ALL cubics have a point of inflection, but the TP cubic is special in that it's the only one with a STATIONARY point of inflection.

Now, you must remember that when you differentiate a function, you get an x-value, not a y-value. So, by differentiating and finding an x-value, you can then use this to find your k. For example, let's use my function as earlier.

1) Differentiate, dy/dx=9x^2-36x+36

2) Solve for zero, x^2-4x+4=0 ===> (x-2)^2=0 ===> x=2

3) Substitute this into the cubic:

f(x)=3x^3-18x^2+36x-21
f(2)=3(2)^3-18(2)^2+36(2)-21=24-72+72-21=3

So, our cubic is of the form f(x)=a(x-2)^3+3. An expansion of this would quickly show that the x^3 has no coefficient, which means a 3 must have been factorised out of the original, so we can put it as f(x)=3(x-2)^3+3, which is exactly what I found earlier with inspection.

[Conic]

The mistake is that expands to , not . Eulerfan's method will work, but he made a slight error (forgot some of the coefficients in the binomial expansion).