VCE Methods Question Thread!

[cosine]

I feel like both these skills are very "practical" based more than theory, so the best way to learn these skills would be through actually talking with a person: ie. Friend, Teacher or Tutor.

However, that might be hard in this period of the year, so what I suggest is search for "Long polynomial division" on google (or youtube) and see what comes up. I think a video tutorial would be the best as it would go through step by step.
Here is one from Khan Academy (usually makes good stuff):
https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-division
And here is a step-by-step text guide:
http://www.purplemath.com/modules/polydiv2.htm

But I think you should do the search and see what clicks with you. However, try your best to meet someone in real life and do it, so you get the skill down patt without any questions or flaws in your working.

Will do! thanks so much!

[Zues]

need help with everything :/

[brightsky]

a) From the diagram, the truncus passes through the points (10,40) and (50,0). After substituting these points into the rule of the truncus, we obtain the following equations.

40 = a/(10)^2 + b......(1)
0 = a/(50)^2 + b......(2)

Solving (1) and (2) simultaneously yields a = 12500/3 and b = -5/3.

b) From the diagram, the domain is [-50, -10] U [10, 50] and the range is [0,40].

c) The wording of this question is quite strange. My assumption is that they are looking for the horizontal distance from the starting point (10,40) to the point along the truncus at which y = 20.

The rule of the truncus is y = 12500/(3x^2) - 5/3. When y = 20:

20 = 12500/(3x^2) - 5/3
x = - 50sqrt(13)/13 or x = 50sqrt(13)/13

Hence, the horizontal distance is (50sqrt(13)/13 - 10) metres.

d) The original truncus has equation y = 12500/(3x^2) - 5/3. After dilation by a factor of 2 from the y-axis, the equation of the image is y = 12500/(3(x/2)^2) - 5/3 = 50000/(3x^2) - 5/3. It is given that the structure remains 40 metres tall.

40 = 50000/(3x^2) - 5/3
x = -20 or x = 20

Hence, the diameter of the new top is 2*20 = 40 metres.

0 = 50000/(3x^2) - 5/3
x = -100 or x = 100

Hence, the diameter of the new base is 2*100 = 200 metres.

It is quite odd for Heinemann to ask a student to comment on how realistic a particular proposal is, especially given that there are no obvious problems with the proposal. However, I'd say that the proposal is probably a little unrealistic given how wide the diameter of the new base is. But, of course, this is a subjective response. Rest assured that you will not get any questions of this sort on an actual VCAA exam.

[Zues]

a) From the diagram, the truncus passes through the points (10,40) and (50,0). After substituting these points into the rule of the truncus, we obtain the following equations.

40 = a/(10)^2 + b......(1)
0 = a/(50)^2 + b......(2)

Solving (1) and (2) simultaneously yields a = 12500/3 and b = -5/3.

b) From the diagram, the domain is [-50, -10] U [10, 50] and the range is [0,40].

c) The wording of this question is quite strange. My assumption is that they are looking for the horizontal distance from the starting point (10,40) to the point along the truncus at which y = 20.

The rule of the truncus is y = 12500/(3x^2) - 5/3. When y = 20:

20 = 12500/(3x^2) - 5/3
x = - 50sqrt(13)/13 or x = 50sqrt(13)/13

Hence, the horizontal distance is (50sqrt(13)/13 - 10) metres.

d) The original truncus has equation y = 12500/(3x^2) - 5/3. After dilation by a factor of 2 from the y-axis, the equation of the image is y = 12500/(3(x/2)^2) - 5/3 = 50000/(3x^2) - 5/3. It is given that the structure remains 40 metres tall.

40 = 50000/(3x^2) - 5/3
x = -20 or x = 20

Hence, the diameter of the new top is 2*20 = 40 metres.

0 = 50000/(3x^2) - 5/3
x = -100 or x = 100

Hence, the diameter of the new base is 2*100 = 200 metres.

It is quite odd for Heinemann to ask a student to comment on how realistic a particular proposal is, especially given that there are no obvious problems with the proposal. However, I'd say that the proposal is probably a little unrealistic given how wide the diameter of the new base is. But, of course, this is a subjective response. Rest assured that you will not get any questions of this sort on an actual VCAA exam.

Thanks for that brightsky

additionally, can someone show me how to graph these, with in depth explanations (if possible) as well, something simple like a will do

[keltingmeith]

So youre telling me to drop synthetic division? I mean it works fine with just finding linear factors of polynomials, but for some odd reason it just wont work out when you need to use it for hyperbolas.

It's not that it won't work, it's just that it has a lot of shortcomings that you've just discovered.

Synthetic division is perfectly fine, and for larger polynomials it is much faster than polynomial division. However, you HAVE to make sure that what you're using to divide is of the form (x-r). THIS is where polynomial division and division by inspection are superior - the denominator can be in ANY form to use either of them.

Note: I wouldn't say that you SHOULDN'T use synthetic division - I even included it in my notes book when I did methods - just make sure you know when it is appropriate to use it, and when using another technique could be faster/easier on you.

[brightsky]

Thanks for that brightsky

additionally, can someone show me how to graph these, with in depth explanations (if possible) as well, something simple like a will do

There are two ways to sketch these kinds of graphs; you can either sketch them using addition of ordinates, or express them as a hybrid function and sketch each component separately over the appropriate domain. I prefer the second method, since addition of ordinates is usually a pain unless you have access to graph paper.

I'll do part a) to elucidate my thought process. The function is y = |x + 4| + |x - 4|.

Case 1: x ≥ 4

Ask yourself: If x is a number greater or equal to 4, after I remove the absolute value sign, do I leave the expression as is or do I multiple the expression by -1? It is clear that if you replace x with a number greater or equal to 4 in |x + 4|, you will always get a positive number or 0 inside the absolute value sign. Hence, if you were to remove the absolute value sign, you would simply leave the expression as is. Similarly, if you replace x with a number greater or equal to 4 in |x - 4|, you will, again, always get a positive number or 0 inside the absolute value sign. Hence, if you were to remove the absolute value sign, you would, again, simply leave the expression as is. The reasoning above accounts for the simplification below:

y = |x + 4| + |x - 4| = (x + 4) + (x - 4) = 2x

Hence, in this domain, the function is simply y = 2x.

Case 2: -4 ≤ x < 4

We repeat the same procedure as above.

y = |x + 4| + |x - 4| = (x + 4) - (x - 4) = 8

Note that after I removed the absolute value sign from |x - 4|, I had to multiply the resultant expression by -1, since if I replaced x with a number in the domain under consideration, I would always obtain a negative number or 0 inside the absolute value sign.

Hence, in this domain, the function is simply y = 8.

Case 3: x < -4

Again, we repeat the same procedure as above.

y = |x + 4| + |x - 4| = -(x + 4) - (x - 4) = -2x

Hence, in this domain, the function is simply y = -2x.

To sketch the graph of y = |x + 4| + |x - 4|, therefore, all you need to do is sketch the graph of y = 2x over the domain x ≥ 4, sketch the graph of y = 8 over the domain -4 ≤ x < 4, and sketch the graph of y = -2x over the domain x < -4.

Hope this makes sense!

[cosine]

It's not that it won't work, it's just that it has a lot of shortcomings that you've just discovered.

Synthetic division is perfectly fine, and for larger polynomials it is much faster than polynomial division. However, you HAVE to make sure that what you're using to divide is of the form (x-r). THIS is where polynomial division and division by inspection are superior - the denominator can be in ANY form to use either of them.

Note: I wouldn't say that you SHOULDN'T use synthetic division - I even included it in my notes book when I did methods - just make sure you know when it is appropriate to use it, and when using another technique could be faster/easier on you.

Could you show me how to do the example I posted please, I mean the denominator is in the form of (x-r) but it still doesnt work -.-

[Zues]

Hi Brightsky, thanks once again. Few points of discussion/queries

• In case 2, why did you minus the two expressions? also can you clarify your explanation here, im working ahead so this is all new to me and i dont quite understand it :/ i understand basic ones like 2|x-3| + 3 for example but not the ... y = |x + 2| + |x − 2| types.
• In case 3, why did you take the negative out, and why again did you minus the two (as apposed to adding it)

Thanks

[Zues]

Ahh i can kinda see

e.g. in case 2

Case 2: -4 ≤ x < 4

We repeat the same procedure as above.

y = |x + 4| + |x - 4| = (x + 4) - (x - 4) = 8

Note that after I removed the absolute value sign from |x - 4|, I had to multiply the resultant expression by -1, since if I replaced x with a number in the domain under consideration, I would always obtain a negative number or 0 inside the absolute value sign.

Hence, in this domain, the function is simply y = 8.

if i subbed -4 into |x-4| i would need to change the sign and because i need to do this i make it negative? if i didnt have to do it, it would plus?

also with case 1/2/3 domains, e.g. 1 in this case.... x ≥ 4   why did you do this, and not x <= 4

also, do you need to do all this for every question? do you take any short cut

Thanks so much

[cosine]

Zeus, dont worry about these questions, I also have Heinemann and my teacher said these questions are not part of the study design, the normal absolute value functions are but not the addition and subtraction of them. So dont stress!

[Zues]

Ahh thats relief then, i now understand it so that sucks to a degree haha

do you know what else is omitted?

[keltingmeith]

Could you show me how to do the example I posted please, I mean the denominator is in the form of (x-r) but it still doesnt work -.-

Okay, so we have - first step, take out that 2, giving us .

Now, we HAVE to do something with that 2. We can't just leave it there, and we can't remove it - so, I'm going to take it out of the fraction for a bit, giving us

Now, we use synthetic division on the , giving us 4 and 9, where 9 is the remainder. So, this means that:

[keltingmeith]

Zeus, dont worry about these questions, I also have Heinemann and my teacher said these questions are not part of the study design, the normal absolute value functions are but not the addition and subtraction of them. So dont stress!

I hate to contradict your teacher, BUT they sort of are covered.

In the study design (page 129 for reference) it says that you need to be able to graph all sums, differences, products and compositions of functions that you can graph individually - including a graph of two modulus functions. (NOTE: you do not have to know how to do quotients - that's specialist material, BUT specialist doesn't assess it directly anyway)

So, while those types of modulus graphs aren't SPECIFICALLY mentioned, they are still assessable. (granted, don't think I've ever seen them before on an exam)

[cosine]

Okay, so we have - first step, take out that 2, giving us .

Now, we HAVE to do something with that 2. We can't just leave it there, and we can't remove it - so, I'm going to take it out of the fraction for a bit, giving us

Now, we use synthetic division on the , giving us 4 and 9, where 9 is the remainder. So, this means that:

Ahh, there we go... its always these careless mistakes....
Thanks heaps though, you are more useful than my teacher, no seriously you are...

[Zues]

hows this 2x?