VCE Methods Question Thread!

[Mehul123]

either 1. find the inverse function by swapping x and y in the original function and isolating y
or 2. reflect the original function over the line y=x

thanks

[Mehul123]

Do you have to be talented to get above 40 in maths methods?

[Shadowxo]

I've found you need a bit of both talent and ability to work hard. If you have more talent, you don't have to work as hard as somebody who doesn't. To get 40+ you don't need to be brilliant but you do need to be willing to put in the time and ability does play a part (in my experience).

[NAT0003]

I've found you need a bit of both talent and ability to work hard. If you have more talent, you don't have to work as hard as somebody who doesn't. To get 40+ you don't need to be brilliant but you do need to be willing to put in the time and ability does play a part (in my experience).

Say if I average 90% in my methods sacs and then get something like 38/40 and 70/80 on the exams. Do you reckon that would get a raw 42

[Cranium002]

A few questions:


#1:

Spoiler

#2:

Spoiler

#3:

Spoiler

#4:

Spoiler

Thanks

Mod Edit: Merged your multiple threads with our main question thread and combined into one post. See PM for reference. Thanks

[deStudent]

Is there any order? People say to go dilations, reflections then translations but the book doesn't do this. Their answers go in random orders?

http://m.imgur.com/a/Yc5t4
For c and f, would my answers be okay? I just stated the transformations occurring on x' and y'.

[AngelWings]

A few questions:


#1:

Spoiler

#2:

Spoiler

#3:

Spoiler

#4:

Spoiler

Thanks

Mod Edit: Merged your multiple threads with our main question thread and combined into one post. See PM for reference. Thanks

I don't have a lot of time right now, so I'll make this more of a guidance post rather than give you a step-by-step guide. Please excuse the lack of LaTeX on here.

Spoiler

Question 1 (14):
Tip 1: Use your discriminant formula (delta = b² - 4ac) and input the appropriate values.
Tip 2: Where discriminant is not less than 0, there will be either (a) 1 solution if discriminant is = 0 or (b) 2 solutions if discriminant is > 0. This should be okay for rational solutions.
This site will be very good to look at if you're not so sure about discriminants.

Question #1 (16):
Tip 3: Check the signs and what the question is asking. Try graphing a few simple examples. How can a + and - occur respectively if it's in a quadratic? What are the signs in the brackets if you factorised said quadratic?

Question #2 (3 k):
Tip 4: Use quadratic formula and make x = 0 to solve. (If you're not too sure check the Solving Quadratic Equations using the Quadratic Formula section on this link.)

Question #3 (1 a):
Tip 5: Use the formula for all quadratics (y = ax² + bx + c). With two unknowns (x and y), you can fill them with your given coordinates (your two x-intercepts).
Tip 6: After using tip 5, complete the square to find the coordinates of the point (vertex).

Question #3 (2 a):
Tip 7: Similar to tips 5 and 6, this question is kind of the reverse of Q#3 (1a). Try using various formulas like complete the square and y = ax² + bx + c (and its factorised forms) to help you in this question.

Question #4 (3 i):
I don't like how I did this one, so I'll leave this one for others to tell you their method.

It's been a while since I did Methods, so I might be just a little rusty here. Please take this with a grain of salt, but I hope I helped a little! (Feel free to tell me if I'm wrong. I'm that rusty.)

[Shadowxo]

A few questions:


#1:

Spoiler

#2:

Spoiler

#3:

Spoiler

#4:

Spoiler

Thanks

Mod Edit: Merged your multiple threads with our main question thread and combined into one post. See PM for reference. Thanks

Lot of questions here
14. The discriminant is

16. It is a positive parabola, u shape, and has a negative y intercept (-c is negative as c is positive)

3k. We just use the quadratic formula. a=2k, b=-4, c=k

You could simplify further but it would be messier.

1. a) We know the turning point, aka vertex, is in the middle of the two x intercepts, as a parabola is symmetrical. So the x coordinate of the vertex is (4+10)/2 = 7
2. Same thing but we need to find the other x intercept. (6+x)/2 = 2, x=-2

3. i)

[Shadowxo]

Is there any order? People say to go dilations, reflections then translations but the book doesn't do this. Their answers go in random orders?

http://m.imgur.com/a/Yc5t4
For c and f, would my answers be okay? I just stated the transformations occurring on x' and y'.

Your answers are fine.
It doesn't matter what order you do the transformations in (unless they specify), but different orders will result in different answers (resulting in the same graph though).
When going from a basic to a complex graph, like c, I prefer to do reflections then dilations then translations, and the opposite direction when going from complex to basic - eg for f) I would have first said "translation of 4 units in the negative direction of the y axis) first, to get rid of the 4. But the order really doesn't matter, it just means your answers will differ from the solutions - just do what feels most right for you

[vcestressed]

Hey
Find the distance between the x int of this equation:
y=x^2-5x+2
this is how i did it but the answer is wrong:  http://imgur.com/a/lcAYu
the correct answer is supposed to be 2 root 17

[Shadowxo]

Hey
Find the distance between the x int of this equation:
y=x^2-5x+2
this is how i did it but the answer is wrong:  http://imgur.com/a/lcAYu
the correct answer is supposed to be 2 root 17

Your answer is correct

[codebreaker1_91]

How do I do 11. B and C?
Thanks!

[Shadowxo]

So for b) and c), the graph you drew in a) will help a lot.
For b): you have to find for y=k, how many x's will give you that k value.
i) You'll see that there are certain y values that are not on the graph. Looking at the graph, you see it goes up in a straight line to (-8,0) and for the second part, x=-8 to x=8, as x gets larger y gets larger (not linearly), until the end of the second part of the function, (8, 4), and then for the third part, it starts at (8,4) and goes down in the shape 1/x, approaching 0.
From this we can see that the range is (-infinity,4], so there is no x value that gets a y value of greater than 4, so for i) We know there are no solutions for f(x)=k where k is (4,infinity)

ii) From the graph we can see there is one solution for f(x)=k for k is (-infinity,0] and k =4

iii) From the graph we can see there are two solutions for f(x)=k for k (0,4)

c) The question reads "find x such that f(x)=1" ( : is such that)
 f(x) is always negative for the first part, so cannot =1  for first part
Second part: x^1/3+2=1, x^1/3=-1, x=-1
Third part: 32/x=1 , x=32
So x = 32, or x=-1.

If you need any clarification just ask

[geminii]

Hi everyone! I was wondering if anyone could help me solve this question:

Consider the simultaneous linear equations:

mx - 6y = 6
4x - my = m

Find the values of m for which the equations have infinitely many solutions.

My teacher said we should put the equation into matrix form, which i have done as follows:

[m   -6]

•   =  [6]
[4   -m]

[y]  =  [m]

My teacher said we then need to use the determinant (ad-bc), which in this case would be:

m^2 - (4)(-6)
= m^2 + 24.

But I am not sure whether the determinant should be less than 0, equal 0 or be greater than 0? I know that if there are no solutions, the determinant should = 0. But what about for infinitely many solutions?

Thank you so much! :D

[deStudent]

http://m.imgur.com/a/Mxg0k

Q5) I'm a bit confused, if when we dilate by a factor in the y-axis, do we apply it to everything inside the function. In this case (x+12), this seems true since you get the correct answer if you do so.

However, if we reflect in the the y-axis would this also mean we multiply (x+12) by -1? My teacher did an example with a similar question, but when he reflected in y, he did not change the sign of the constant inside of the function, he only did it to x.

Q4f) I did this both algebraically and by observation, however if I do it by observation it contradicts what I wrote in the image. The problem arises (kind of relates to the problem in Q5) after I translate in x and dilate. If I reflect in y (as in multiply entire inside function by -1), I get x-3. Wouldn't this suggest that I would have to translate 3 units negatively in the x-axis instead of positive (as seen in the image)?

Cheers