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Higher Maths 2013-2014 : Discussion and Help Thread

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Multiply your brackets again carefully..
I find writing them like this is easier when squaring them.
(x^2 + 2) (x^2 + 2)

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Original post by Pennyarcade
Multiply your brackets again carefully..
I find writing them like this is easier when squaring them.
(x^2 + 2) (x^2 + 2)

Posted from TSR Mobile


Oh... I totally forgot how to square brackets :tongue: Thanks again
Reply 22
Avatar for I am Ace
I am Ace
OP
4
DLB Maths has just put out a unit 1 Outcome 1 practice NAB video

http://www.youtube.com/watch?v=Tuz60mRUhic
(edited 10 years ago)
Subbing. Doesn't look like the best of courses, but I hope it should be pretty fun regardless.
Reply 24
Avatar for Memetics
Memetics
8
Just had a talk with someone who might be my maths tutor this year(S5), so that's an interesting thing for me. Never had one.
Anyone on here ever had one? Are they effective?:rolleyes:
Original post by Memetics
Just had a talk with someone who might be my maths tutor this year(S5), so that's an interesting thing for me. Never had one.
Anyone on here ever had one? Are they effective?:rolleyes:


I got one at about Christmas time last year for Higher Maths and he helped a lot :biggrin:(Band 1 A :cool:) But beware as some tutors can be a waste of time and money. Look at there reviews and try to get someone who is young as they have past experiences with the higher course and possibly have recently completed or are studying towards a degree requiring the use of maths. They way I did it was have him go through the HSN notes and try examples from there then towards the exam he made up exam 'like' questions and helped me with any past paper questions I had a problem with. But my friend had a tutor and just did past paper questions all year long... (they got a B and the guy must have been like 70 years old:eek:, they were perfectly capable of getting an A if they worked on understanding the maths more rather than just plowing through questions, I often found myself explaining concepts to them in our study periods.)
(edited 10 years ago)
Hi guys, wondering if I could get a bit of help.

im doing the following question:
Find the stationary points and determine the nature of curve y=(x-2)(x-3)^2

I really don't get how to do this question at all. I know the theory behind it but I can't put it into practice for this question. Could somebody please help? Most appreciated :smile: (I tend to use a 2nd derivative rather than a nature table for determining the nature if that's useful at all?)
Original post by jackamakka
Hi guys, wondering if I could get a bit of help.

im doing the following question:
Find the stationary points and determine the nature of curve y=(x-2)(x-3)^2

I really don't get how to do this question at all. I know the theory behind it but I can't put it into practice for this question. Could somebody please help? Most appreciated :smile: (I tend to use a 2nd derivative rather than a nature table for determining the nature if that's useful at all?)


Why don't you show us your working, or what you've tried first? How are we meant to know where you're stuck if you just give us the question and say you're stuck!? :tongue:
I have no working as I dont have a clue what to do ..
The diagram shows a sketch of te graphs of the quadratic functions
Y=f(x) and
Y=-f(x + 1/2)
A) if f(x) has a stationary value of (1,0) find f(x)
Thanks guys!

Posted from TSR Mobile
1378225046428.jpg36.3KB
Original post by Hype en Ecosse
Why don't you show us your working, or what you've tried first? How are we meant to know where you're stuck if you just give us the question and say you're stuck!? :tongue:


Well, I was reattempting it as I didn't really know how to go about it.

The first time I got that a Stationary value occurs when dy/dx = 0

So, to get the equation to a point where I could differentiate I thought to break the brackets:

y=(x^2 + x - 6)^2
y=(x^2 + x - 6)(x^2 + x - 6)
y=x^4 + x^3 - 6x^2 + x^3 + x^2 - 6x - 6x^2 - 6x + 36
y=x^4 + 2x^3 - 11x^2 - 12x - 36

So then, I just differentiated that:
dy/dx = 4x^3 + 6x^2 - 22x - 12

Equate to 0 and factorise
0 = 2(2x^3 + 3x^2 - 11x - 6)

I can't think of how to factorise that further so as to find the roots of the equation (the x = something). I don't know if I've made a mistake or whatever, but it just looks wrong... :frown:
Original post by Pennyarcade
I have no working as I dont have a clue what to do ..
The diagram shows a sketch of te graphs of the quadratic functions
Y=f(x) and
Y=-f(x + 1/2)
A) if f(x) has a stationary value of (1,0) find f(x)
Thanks guys!

Posted from TSR Mobile


Oh, this is a fun one! To start you off...what's the general equation for a quadratic?

Original post by jackamakka
Well, I was reattempting it as I didn't really know how to go about it.

The first time I got that a Stationary value occurs when dy/dx = 0

So, to get the equation to a point where I could differentiate I thought to break the brackets:

y = (x - 2)(x - 3)^2
y=(x^2 + x - 6)^2
y=(x^2 + x - 6)(x^2 + x - 6)
y=x^4 + x^3 - 6x^2 + x^3 + x^2 - 6x - 6x^2 - 6x + 36
y=x^4 + 2x^3 - 11x^2 - 12x - 36

So then, I just differentiated that:
dy/dx = 4x^3 + 6x^2 - 22x - 12

Equate to 0 and factorise
0 = 2(2x^3 + 3x^2 - 11x - 6)

I can't think of how to factorise that further so as to find the roots of the equation (the x = something). I don't know if I've made a mistake or whatever, but it just looks wrong... :frown:


You're following the right process, but you've made a silly mistake that you're going to kick yourself for. I've bolded it for you there. Are you sure that's how you expand the brackets correctly? What you've got there is an expression of the form xy2xy^2, you turned that into (xy)2(xy)^2. Can you see why you can't do that? :tongue:
(edited 10 years ago)
Reply 31
Avatar for I am Ace
I am Ace
OP
4
Original post by jackamakka
Well, I was reattempting it as I didn't really know how to go about it.

The first time I got that a Stationary value occurs when dy/dx = 0

So, to get the equation to a point where I could differentiate I thought to break the brackets:

y=(x^2 + x - 6)^2
y=(x^2 + x - 6)(x^2 + x - 6)
y=x^4 + x^3 - 6x^2 + x^3 + x^2 - 6x - 6x^2 - 6x + 36
y=x^4 + 2x^3 - 11x^2 - 12x - 36

So then, I just differentiated that:
dy/dx = 4x^3 + 6x^2 - 22x - 12

Equate to 0 and factorise
0 = 2(2x^3 + 3x^2 - 11x - 6)

I can't think of how to factorise that further so as to find the roots of the equation (the x = something). I don't know if I've made a mistake or whatever, but it just looks wrong... :frown:


This is a chain rule question, it is taught in the Unit "Further Calculus - Differentiation." I won't imagine you have covered this, I haven't looked back to the source. Is this past paper or did your teacher give you it?
Original post by Hype en Ecosse
Oh, this is a fun one! To start you off...what's the general equation for a quadratic?


Ax^2 +bx + c = 0
Now what :P?
Original post by Hype en Ecosse

You're following the right process, but you've made a silly mistake that you're going to kick yourself for. I've bolded it for you there. Are you sure that's how you expand the brackets correctly? What you've got there is an expression of the form xy2xy^2, you turned that into (xy)2(xy)^2. Can you see why you can't do that? :tongue:


Ah yeah, thanks! Sorry, I didn't realise that when I was doing the question. Thank you :smile:

Original post by I am Ace
This is a chain rule question, it is taught in the Unit "Further Calculus - Differentiation." I won't imagine you have covered this, I haven't looked back to the source. Is this past paper or did your teacher give you it?


Yeah, we've been doing that Unit at the moment. Not heard the term chain rule though? My teacher gave it to us for homework
Original post by I am Ace
This is a chain rule question, it is taught in the Unit "Further Calculus - Differentiation." I won't imagine you have covered this, I haven't looked back to the source. Is this past paper or did your teacher give you it?


You don't need to use the chain rule, bud. :tongue:

Original post by Pennyarcade
Ax^2 +bx + c = 0
Now what :P?


So you know that your graph that you're trying to find, f(x), has that form. So you basically have to figure out the values of a, b and c, don't you? Think you can try to do that?
Original post by Hype en Ecosse


So you know that your graph that you're trying to find, f(x), has that form. So you basically have to figure out the values of a, b and c, don't you? Think you can try to do that?


Ah right, thanks :biggrin:
Write tthe trignometric function represented by each graph.
I know that its going to be
Y= sin (x-SOMETHING) - 2
I dont really know how to find that something though.
I assume it might have something to do with it reaching its highest point after 180 instead of 90 though..

Posted from TSR Mobile
1378239893709.jpg37.8KB
Original post by Pennyarcade
Write tthe trignometric function represented by each graph.
I know that its going to be
Y= sin (x-SOMETHING) - 2
I dont really know how to find that something though.
I assume it might have something to do with it reaching its highest point after 180 instead of 90 though..

Posted from TSR Mobile


We've done this question in class today. But our teacher assumed it was a Cos graph so I'm not sure how to help.

But assuming it's a Sin graph then the highest point should be at 90. The highest point is somewhere between 180 and 270 (225 - random guess :wink: )

So just do 225 - 90 = 135
Y = sin (x-135) - 2

( Or if it's a cos graph Y = - cos (x - 45) -2 )
Original post by Nessie162
We've done this question in class today. But our teacher assumed it was a Cos graph so I'm not sure how to help.

But assuming it's a Sin graph then the highest point should be at 90. The highest point is somewhere between 180 and 270 (225 - random guess :wink: )

So just do 225 - 90 = 135
Y = sin (x-135) - 2

( Or if it's a cos graph Y = - cos (x - 45) -2 )


But the answer said it is Y = sin(x-120) - 2 :frown:
Original post by Nessie162
We've done this question in class today. But our teacher assumed it was a Cos graph so I'm not sure how to help.

But assuming it's a Sin graph then the highest point should be at 90. The highest point is somewhere between 180 and 270 (225 - random guess :wink: )

So just do 225 - 90 = 135
Y = sin (x-135) - 2

( Or if it's a cos graph Y = - cos (x - 45) -2 )


Sin graphs and cos graphs are the same, just one's shifted across a bit relative to the other. sin⁡θ=cos⁡(90−θ)\sin \theta = \cos (90 - \theta) and cos⁡θ=sin⁡(90−θ)\cos \theta = \sin (90 - \theta)

Original post by Pennyarcade
But the answer said it is Y = sin(x-120) - 2 :frown:


Like she said, you're just eyeballing where the maximum is (or any other significant point on the graph that you fancy using for reference). She decided that it looked about halfway between. I'd say it looks a bit less than half and would've went with a nice round 30, which gets you the right answer.
(edited 10 years ago)

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