VCE Methods Question Thread!

[abcdqdxD]

Is [-6,6] the same as {x:x≥-6}U{x:x≤6}?

Also, in exams, how do assessors expect domain/ranges to be shown?

i.e. domain: [0,oo) can be written as {x:x≥0}, Domain: R+U{0}, which one would they prefer?

Thanks

[TrueTears]

Is [-6,6] the same as {x:x≥-6}U{x:x≤6}?

change it to an intersection and it's the same

[abcdqdxD]

change it to an intersection and it's the same

Oh ok thanks!

[BubbleWrapMan]

Is [-6,6] the same as {x:x≥-6}U{x:x≤6}?

Also, in exams, how do assessors expect domain/ranges to be shown?

i.e. domain: [0,oo) can be written as {x:x≥0}, Domain: R+U{0}, which one would they prefer?

Thanks

If it's correct it will be marked correct, unless they ask for a specific form, which they usually don't

[abcdqdxD]

If it's correct it will be marked correct, unless they ask for a specific form, which they usually don't

What are the names of the forms? And which ones do they refer to respectively?

[BubbleWrapMan]

They don't have names, they would say something like "in the form [a,∞)", though for sets they usually don't do that

[KevinooBz]

What are the names of the forms? And which ones do they refer to respectively?

Do you mean like set and interval notation?

[vashappenin]

Hey can someone please help me with these questions?
Determine the set of solutions of the simultaneous system



Find the values of c,d and e such that the following simultaneous equation has no solution
I know there's the method where you group each letter into fractions or something but I really don't like that method.. Isn't there another method where you find the determinant and then sub in some numbers and get the answers?? :S

Also, how come it says the equation has no solutions yet we still get solutions in the end??


And can someone please explain to me how to solve sets of 3 simultaneous Equations easily?? I always struggle to solve them!!

Thanks

[e^1]

Find the values of c,d and e such that the following simultaneous equation has no solution
I know there's the method where you group each letter into fractions or something but I really don't like that method.. Isn't there another method where you find the determinant and then sub in some numbers and get the answers?? :S

Attempt the question before you see it!

For the two lines to have no solutions, then the lines must never intersect with each other. To accomplish this, for instance, if both were in the form of , then we make both lines equal in gradient, but vertical translation must never equal so that they never intersect.

In this case, we can use matrices and find the determinant, then make it equal to zero to find the value of one variable that will make the system of 2 linear equations have either infinite solutions, or no solutions.

Firstly, divide the second equation by half so that the both of the equations have the coefficient of x as equal. For the left side of the equation, we can use the matrix method explained above.

Note that this is one way of approaching of finding the value of c, and you can do this without using the matrix method (you can find that for yourself, it's also in front of you).



Great! We've fulfilled the first requirement of having a system of linear equations of no solutions.

Now to deal with the constants at the right side of both equations (assuming z is an arbitrary constant, as x and y are independent and dependent variables respectively). Both lines must not have the same vertical translations, hence:



Just to make it clear for the and/or bit:
(In English: e can equal 4 but d must not equal -6) OR
OR


These are the possible solutions for d and e with , for the system of linear equations to have no solution.

If you don't understand this last bit

To solve this, imagine them both as linear equations. If both lines are equal (infinite solution), then the substituting those values of d and e will make the system of linear equations behave as having infinite solutions.

I'll leave it there, hopefully you can finish the rest from here

Also, how come it says the equation has no solutions yet we still get solutions in the end??

The question is asking you to find the selected variables so that the system of linear equations has no solutions (ie. Each linear equation of the system never intersects the other two.)

Determine the set of solutions of the simultaneous system

...

Transform it into matrix form:



Multiply the inverse of     on both sides of the equation, using CAS calculator or a similar device.

But what eventually happens is that it gives an error, that by inversing it gives a singular matrix. What this means is that the system of linear equations have no solution, or have infinite solutions. In this case it is a no solution, because the three linear equations are not identical to each other.

Hence, there are no solutions for this simultaneous system.

And can someone please explain to me how to solve sets of 3 simultaneous Equations easily?? I always struggle to solve them!!

For three equations with three unknowns:

http://www.themathpage.com/alg/simultaneous-equations-3.htm (manual way)
Re: Methods [3/4] Question Thread!


Once again, please ask if you don't understand something (or if I did something wrong). These questions are also new to me too

[polar]

OMG THIS IS SO COOL

nice find, e^1

[vashappenin]

Heu thanks! I probs shpuldv mentioned this earlier, but i'm expected to dp these by hand, so can you please help me do it by hand?? Thanks

[e^1]

Heu thanks! I probs shpuldv mentioned this earlier, but i'm expected to dp these by hand, so can you please help me do it by hand?? Thanks

Answer: Re: Methods [3/4] Question Thread!

No, you will usually deal with 2 simultaneous equations, as to dealing with "no, infinite or unique solutions" questions.
The same goes with doing a 3x3 matrix, you can do that by CAS.

If you wanted to do the second question by hand, work it out by elimination and substitution, using all equations.

More importantly, you should understand the idea of what a no, unique or infinite solution is, so that you can do these types of questions.

[abcdqdxD]

Can someone give some tips for graphing quartics with only x intercepts and no stationery points.I find it quite annoying because we don't know the turning points.

And a lot of quartics look different because of their different number of x intercepts, so can someone show a sample equation of each possible form of quartic, and how its graph would look? e.g. y=(x+2)^2 (x-3)(x-1); y=(x+2)^3(x+1); etc.

And in the case of (a+b)^n, why is it only odd number values of 'n' have stationery points (3,5,7,9, etc.), and n=1 'cuts' while n=2 (or any even number) 'touches' the axis?

Thanks

[b^3]

When you have the equation in factorised form, the power on each bracketed term will tell you what kind of intercept you have.
i.e.
cuts the axis at (as you've said)
will have a turning point on the intercept at (note you've said this isn't a stationary point as it just 'touches the axis', it is a stationary point because the gradient at that point is zero.
will have a stationary point of inflection at
will have a turning point, just like except the gradient change on either side will be steeper for  at .
...and so on..

When you go to draw something like , we can already tell that the graph will cut at , and have a turning point that just touches at . Now you also need to know whether the graph is flipped in the axis or not, in this case since there is no negative out the front of the brackets and all the coefficients on inside the brackets are positive, we know that it will be the non flipped version of the cubic.

https://www.desmos.com/calculator/gfnb4uzp2l

Can someone give some tips for graphing quartics with only x intercepts and no stationery points.I find it quite annoying because we don't know the turning points.

You probably haven't covered calculus yet so far this year, although you may have covered it in year 11. Not sure if there is a proper method for it pre-calculus, but anyways, you'd find the derivative, let it equal zero (since this will be where the stationary points are) and then solve for the values of x. Then from that find a y value for each x.

The only other way I can think of is that if you know that you have three x intercepts for a quartic, you also know that if the graph starts from the top then it will leave towards the top too (or starts from the bottom then leaves from the bottom too), as with the other even power graphs. Then you would see that two intercepts can only cut while the third would have to touch (i.e. the cutting intercepts are power one, while the touching intercept would be power two, 1+1+2=4), but to get back to the next intercept, you'd need turning points in between the intercepts. If you only had two intercepts then you could have two different situations, one would have to have both intercepts just touching (power 2, 2+2=4). The second could have a stationary point of inflection intercept and one intercept that cuts (3+1=4).

It might be a bit of an idea to play around on that online graphics calc I linked

Hope that helps

[darklight]

Keeping up with the matrices trend:
2F Essentials - attached.

How does one complete part b?