VCE Methods Question Thread!

[knightrider]

For this question

find Q(x) and R(x) when is divided by

I got that and

The answers in the book agree with me on Q(x) but for R(x) the book got

Am i right or is the book right?

[IndefatigableLover]

I was just wondering,
Are we expected to solve this by hand? (it involves solving for 4 unknowns)

Find the equation of the quartic function for which the graph passes through the points with coordinates:
(-1,43) , (0,40), (2,70), (6, 1618), (10,670)

Even if we don't need to know, could someone please solve it for me? I tried several times and got some really bizarre answers

^^
You'll never get a question like that on a VCAA exam, unless it's with a calculator.

Yep that's definitely true!
Though SE_JM, I'll show you a relatively unknown method on solving for these types of questions (that is if you use a TI-Nspire) since you want to be able to fully utilise your calculator skills as efficiently as possible (typing out simultaneous equations can be time consuming).

Basically go on the 'Main Menu', head to 'Vernier DataQuest' and towards the bottom left you'll see three buttons (from left to right it's a timer, graph and data table). Click on the data table and then from ascending order in 'x' values, input your coordinates in (there's separate columns for 'x' and 'y').

Once when you're done, click on the graph button on the bottom right and you'll be taken to a graph. Now if you hit [Menu] and go to Analyse [4] and then Curve Fit [6] and fit it to a quartic [4], you'll get your coefficient values for a, b, c, d, and e which you would sub into the general equation for a quartic (which EulerFan101 has stated above).

For this question

find Q(x) and R(x) when is divided by

I got that and

The answers in the book agree with me on Q(x) but for R(x) the book got

Am i right or is the book right?

It's the same thing except yours is in Mixed fraction form and there's is in improper fraction form.

[knightrider]

It's the same
 thing except yours is in Mixed fraction form and there's is in improper fraction form.

Thanks IndefatigableLover 

But i dont get it.
when you expand their mixed fraction    this gives us=

So what is wrong?

[IndefatigableLover]

Thanks IndefatigableLover 

But i dont get it.
when you expand their mixed fraction    this gives us=

So what is wrong?

EDIT: It is correct. The 'negative' part is for overall is you would do just -[(3x27) + 20]/27
Basically it's the same answer.

[knightrider]

EDIT: It is correct. The 'negative' part is for overall is you would do just -[(3x27) + 20]/27
Basically it's the same answer.

Yep Thanks IndefatigableLover

so whenever you would have a negative it is for the overall part right?

[IndefatigableLover]

Yep Thanks IndefatigableLover

so whenever you would have a negative it is for the overall part right?

Yep it's for the overall part!
I guess one way to double check when in times of doubt is to chuck in the improper fraction into a normal scientific calculator and change it into a mixed fraction to see if it's the same thing

[Zues]

I know with absolute function (modulus) graphs if you have say | f(x)|, this flips the entire section of the graph below the x axis above it. And if its f(|x|) then this causes the right side of the graph to flip over and replace the left.

So if i have y = 2|sinx| - 3 how do i go about it? neither the whole function is "modulated" nor is each individual x variable. Say i had y = |x^2| - 7, would i just follow that id draw the graph within the modulus and then because its |x| variable i flip it over (right over left) then shift 7 units down? so in the case with sin, id draw the sin graph then flip negative y's to make positive, then apply the amplitude of 2 and shift 3 down? but if i had y = |2sin(x) - 3| then id graph the whole entire thing then do the reflection in x axis to positive y's? i hope that makes sense.

im a bit confused because say it asks me to sketch y = 2|x-3| + 1, i know i can set two hybrid functions being y = 2x - 5 for x>= 3 and y= -2x + 7 for x<3, but how do i sketch this. it says one thing for x>=3 and x<3 but the x intercepts for each graph is 5/2 and 7/2 respectively? Hopefully that makes sense..

Thanks

[keltingmeith]

I know with absolute function (modulus) graphs if you have say | f(x)|, this flips the entire section of the graph below the x axis above it. And if its f(|x|) then this causes the right side of the graph to flip over and replace the left.

So if i have y = 2|sinx| - 3 how do i go about it? neither the whole function is "modulated" nor is each individual x variable. Say i had y = |x^2| - 7, would i just follow that id draw the graph within the modulus and then because its |x| variable i flip it over (right over left) then shift 7 units down? so in the case with sin, id draw the sin graph then flip negative y's to make positive, then apply the amplitude of 2 and shift 3 down? but if i had y = |2sin(x) - 3| then id graph the whole entire thing then do the reflection in x axis to positive y's? i hope that makes sense.

The same way you go about drawing any graph - break it down to simplest form, then slowly apply transformations until you've the graph you want.

1. Identify the simplest graph, y=sin(x)
2. Start applying transformations. I'd go like this:
y=sin(x) ---> y=|sin(x)| ---> y=2|sin(x)| ---> y=2|sin(x)|-3
3. Using the above steps, draw the graph at each stage (even just rough sketches), since one transformation at a time is much easier than several.

So, first you'd sketch y=sin(x), then you'd apply the transformation |f(x)| to it. The new graph, you would then dilate by factor 2 from the x-axis. Finally, you'd translate all of that down 3.

Also - |x^2|=x^2. So, your second example is incredibly easy to draw.

im a bit confused because say it asks me to sketch y = 2|x-3| + 1, i know i can set two hybrid functions being y = 2x - 5 for x>= 3 and y= -2x + 7 for x<3, but how do i sketch this. it says one thing for x>=3 and x<3 but the x intercepts for each graph is 5/2 and 7/2 respectively? Hopefully that makes sense..

Thanks

I'm confused as to why this is an issue? For x<3, we graph y=7-2x, but it has an x-intercept >3, and so it wouldn't appear on the graph. For x>3, we graph y=2x-5, which has an intercept <3, so it also wouldn't appear on the graph. So, there is no conflicting point.

In fact, these two graphs should only share one point, which is the vertex of the modulus, (3, 1). It makes sense that the other points (such as intercepts) aren't the same.

[Zues]

that makes sense now since if i continued the graph past (3,1) it would pass through each respective x intercepts.

with these
c) how did they get x - 1 + 1, where did the -1 come from
d2)  2nd to third line, where did the e^loge go?

[Conic]

For , you should be aware that , as is the inverse of . In the first question you have and for the second, .

[knightrider]

I just tried factorising a particular cubic and just managed to factorise it using a method.

i wanted to know if this method was a proper method and can be applied to other cubics factorised this same way or was this just a fluke.

 

[keltingmeith]

I just tried factorising a particular cubic and just managed to factorise it using a method.

i wanted to know if this method was a proper method and can be applied to other cubics factorised this same way or was this just a fluke.

 

This method can be applied to any polynomial - personally, I find that often it's hard to see this method, whereas division is a lot more concrete, but if you want to try it, go for it.

[Phy124]

I just tried factorising a particular cubic and just managed to factorise it using a method.

i wanted to know if this method was a proper method and can be applied to other cubics factorised this same way or was this just a fluke.

 

I believe the method is called "factoring by grouping" if you wish to do a little bit more research and see if it will be a viable method for you to use in the future.

[Splash-Tackle-Flail]

In "Modelling", when we plot the data points after transforming the x scale, the book says to draw a line of best fit to find the gradient and check for vertical translation- is this done by eye or do we do a regression line in CAS?

[keltingmeith]

In "Modelling", when we plot the data points after transforming the x scale, the book says to draw a line of best fit to find the gradient and check for vertical translation- is this done by eye or do we do a regression line in CAS?

... Do you mean to post this on the Further thread, or is your textbook doing something weird?