VCE Methods Question Thread!

[polar]

the second one is the 'usual' way you transform functions in exams - something 'basic' to something more 'complex', in the first one, you're going from something 'complex' to something 'basic'

don't look if you want to save the exam for the end of the year

to
what do I substitute into that x to go from f to g?

hence, also, and you just represent that as a matrix

[Will T]

Essentials chapter 3 chapter review extended response question 6.....

I can not get through it, in dire need of some assistance.

[kenners]

the second one is the 'usual' way you transform functions in exams - something 'basic' to something more 'complex', in the first one, you're going from something 'complex' to something 'basic'

don't look if you want to save the exam for the end of the year

to
what do I substitute into that x to go from f to g?

hence, also, and you just represent that as a matrix

for the first graph though, wouldnt dilating by a factor of 2 from y axis affect the horizontal translation? so 1/2x+4 become 1/x+2?
therfore only a translation of 2 required to achieve the more simple graph
my brain hurts

[polar]

yeah, you can do it that way, but remember to also multiply the 3 next to it

actually, if you did it this way, the answer wouldn't be in the form required and you'd need to manipulate it a bit

[kenners]

still confused as to why the first graph has dilation factor of 2 from the y-axis and a horizontal translation of 4 while the second graph is dilation factor of 1/2 and horizontal translation of 2
as they are opposites, shouldnt the first graph have a horizontal translation of 8
Graph 1: (8 * factor of 1/2 will give the translation of 4 as required by the graph)
Graph 2:(2 * factor of 2 will give the translation of 4 as required by the graph)
however graph 1 only states transformation of 4

im so confused ):

it just seems to me like although these are graphs with opposite transformations, the matrix transformations required are not opposites

[polar]

for this question, they needed you to write it in a specific form, so it's easier to do it mathematically than visually

it just seems to me like although these are graphs with opposite transformations, the matrix transformations required are not opposites

Spoiler

the transformation in the second one is

solving for gives which is the matrix for graph 1. so they are opposites of each other
so that's another way of doing the question, just work out the translation the other way and then solve backwards.

[kenners]

for this question, they needed you to write it in a specific form, so it's easier to do it mathematically than visually

Spoiler

the transformation in the second one is

solving for gives which is the matrix for graph 1. so they are opposites of each other
so that's another way of doing the question, just work out the translation the other way and then solve backwards.

Thanks!!! finally get it haha, things get confusing when its done the other direction

[Planck's constant]

Essentials chapter 3 chapter review extended response question 6.....

I can not get through it, in dire need of some assistance.

WillT,

Here's 6ai (rest by analogy)

y=f(x)    (1)

The sequence of trnsformations :

*a translation of 3 units in the positive direction of x-axis
*a translation of 5 units in the positive direction of the y-axis
*a reflection in the line y = x

means

(x, y) -> (x+3, y+5) -> (y+5, x+3) = (x', y')

Therefore,

x' = y+5    (2)
y' = x+3    (3)

Solve (2), (3) for x, y

y=x'-5   (4)
x=y'-3    (5)

Substitute (4), (5) into (1)

x'-5 = f(y'-3)

apply f-1 both sides:

f-1(x'-5) = f-1f(y'-3)
f-1(x'-5) = y'-3
y'-3 = f-1(x'-5)   --- and after pissing off the dashes

y = f-1(x-5)  + 3

[Will T]

Thanks very much, I did manage to get it eventually but you highlighted another process for handling it. Cheers.

[awesomejames]

Desperate help with this differentiation extended response question!

Let f(x) = ax^2 +bx  where a and b are constants and g(x)= 5x^2 +18

A) find in terms of a and b the gradient of the curve y=f(x) at x=-3

B) find the gradient of the curve y=g(x) at x=-3

C) if the tangent lines to the curve y=f(x) and y=g(x) are parallel at x=-3:
i find an equation relating a and b
iihence find the values of a and b, given that the curves y=f(x) and y=g(x) in fact touch at x=-3


So guys I'm mainly having trouble with part c) i and ii
How would you approach this question?
Thanks in advance

[clueless123]

C)i. Since the tangent lines to the curve are parallel at x=-3, you know their gradients must be equal. (obtain from part A and B)
ii. If they both touch at x =-3, they must also have the same y-co-ordinate. (sub into f(x) and g(x))
You obtain one equation from both (i) and (ii) then solve

[awesomejames]

C)i. Since the tangent lines to the curve are parallel at x=-3, you know their gradients must be equal. (obtain from part A and B)
ii. If they both touch at x =-3, they must also have the same y-co-ordinate. (sub into f(x) and g(x))
You obtain one equation from both (i) and (ii) then solve

But for c)I the equations I got were f'(-3) = -6a + b and g'(-3)= -30

So how does that work? What do I do? Would it be -6a + b = -30

I don't understand :/

Bit more explanation would help

[clueless123]

f'(-3) = -6a + b tells you that the gradient of f(x) at the point x = -3 is "-6a + b"
Because you know that the gradients of the two graphs are the same at x=-3, you know that -6a+b must equal -30(the gradient of g(x) at that point), as you've worked out. That should be your answer to part (i).
For C)ii), you are unable to solve for a & b as there are two unknowns present, with only 1 equation. To fix that, they tell you that the two graphs also touch at the point x=-3, meaning, at that point, the y coordinate is the same.
g(-3) = 63 --> at the point x=-3 the y-coordinate of g(x) = 63.
f(-3) = 9a-3b --> at the point x = -3 the y-coordinate of f(x) is given by 9a-3b.
Because you know that the y-coordinate for both graphs are the same, 63=9a-3b. That's your 2nd equation.

[sin0001]

Alrite so you've already got the equation from part ci) -6a+ b = -30
But part cii) also states that the graphs share the same co-ordinates at x=-3
Therefore, f(-3) = g(-3), so 9a-3b = 63
Now you have two equations, just solve for a & b
Edit: beaten

[b^3]

I may be overthinking this a little, but just want to make the distinction that if the two curves are 'touching', then yes they have the same y coordinate, but just because two curves have the same y coordinate does not mean that they are necessarily 'touching', the could cross each other. In this case since we only get one set of solutions, it fits and works -> https://www.desmos.com/calculator/gbbikm8o0y . They way I initially did it was to let the two equations equal each other in terms of x (yes y coordinates are the same), and then found the discriminant, let it equal zero so that the two curves only had one intersection point. It's more working in this case... but I think what I'm trying to say is that you should be making the connection that 'touching' means that the curves don't cross, they only have one solution, solving for the y values being equal may not always work, it may cause problems later on. It really depends on what set of equations you get.

....Anyways overthinking it, but thought I'd point it out anyways. It works for this question