VCE Methods Question Thread!

[BubbleWrapMan]

[Homer]

Find the values of p and q for which the equations
-3x + py = q
4x - 5y = 20        have,

i) a unique solution
ii) no solution
iii) infinitely many solutions.

[b^3]

There is a couple of ways to do this. (i.e. determinate methods or rearranging e.t.c)

Using the rearranging method we get.
   and
.

For a unique solution we need the lines to be two different lines and intersecting. That is the gradients cannot be the same (if they were they'd be parallel and wouldn't intersect). So you would have
Giving

For no soultion, the lines have to not intersect, that is the gradients have to be the same but the y-intercepts have to be different (if the y-intercepts were the same then you'd have the same line, i.e. infinite sols).
So you can equate the gradients and make the y-ints each other.

For infinite sols, we need the gradients to be the same and the y-intercepts to be the same, so that we get the same lines.

You should be able to get it from that

[Homer]

You made it crystal clear! Thank you so much

[Homer]

The graph of the function with the rule y = x^(2/3) is transformed as follows:

A dilation by a factor of 2 from the y-axis
A reflection in the x-axis
A translation of +4 units parallel to the x-axis
A translation of +1 units parallel to the y-axis

Write down the equation of the rule of the transformed function.

[b^3]

Firstly find the mapping of the transformations.

Then find and in terms of their images and . (since )

Then substitute and into the original function that we are transforming.

[Homer]

I don't understand these transformation.
A dilation by a factor of 2 from the y-axis
A reflection in the x-axis

If it is dilation from 2 from the y-axis why is it (2x,y) instead of (x,2y) and for the reflection why isn't it (-x,y) instead of (x,-y)

[b^3]

Think about it this way, a dilation by factor of 2 from the y-axis is going to be streching it away from the y-axis, that is the x-values will be changing.

For the reflection, if we reflect in the x-axis, we are flipping over/across it/mirroring in the x-axis. So that is the x-values aren't the ones being altered, its the y-values that are being changed, in this case all the positive y values become negative and all the negative y values become positive.

[Homer]

Photo needs to be uploaded again

[nisha]

Hey guys. I don't understand the second step of the worked solution. Any help would be appreciated.

Question 2. where f(x)=x(x-2)^3 +2
2a. If f ′(x) = (ax + b)(x − 2)^2 , where a and b are constants, use calculus to find the values
of a and b. Worked solution:::
Using the product rule gives
f′(x)=3x(x−2)^2 +(x−2)^3
= (x − 2)^2 (3x + x − 2)
=(x−2)^2(4x−2) so a=4,b=−2

EDIT: Don't worry. I got it.

[paulsterio]

Hey guys. I don't understand the second step of the worked solution. Any help would be appreciated.

Question 2. where f(x)=x(x-2)^3 +2
2a. If f ′(x) = (ax + b)(x − 2)^2 , where a and b are constants, use calculus to find the values
of a and b. Worked solution:::
Using the product rule gives
f′(x)=3x(x−2)^2 +(x−2)^3
= (x − 2)^2 (3x + x − 2)
=(x−2)^2(4x−2) so a=4,b=−2

It's a simple factorisation, let's just say let

So

[Tonychet2]

how do u rewrite (1/2x+1) - 1 in the form ax/bx+c ??


I know how to get from the 2nd to the first but ive never learnt how to do the reverse!

p.s. how do i write with mathematical notation on this forum lol

[b^3]

p.s. how do i write with mathematical notation on this forum lol

Use this to generate the code (after a while you'll learn it without needing it) -> http://www.codecogs.com/latex/eqneditor.php
Then you put that in 'tex' tags
e.g.

Code: [Select]

[tex]b^{3}[/tex]
Gives

Code: [Select]

[tex]\int 3b^{2}dx=b^{3}+C[/tex]


[Tonychet2]

[WhoTookMyUsername]

Is this b quarticised?