Author Topic: C1 and C2 DOUBTS HERE!!!!!  (Read 88593 times)

Offline Blizz_rb93

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Re: C1 DOUBTS HERE!!!!!
« Reply #240 on: September 07, 2010, 10:39:39 am »
thank you!
i also would like to ask another small question

now when we have an equation y=6-10x-4x^2 and we move the values to the normal way like putting 4x^2 in the beginning then followed by -10x then by -6, does moving the values affect the signs?? if so, how?

Offline astarmathsandphysics

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Re: C1 DOUBTS HERE!!!!!
« Reply #241 on: September 07, 2010, 02:15:15 pm »
reflection in x axis. All the y vlues change sign

Offline Blizz_rb93

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Re: C1 DOUBTS HERE!!!!!
« Reply #242 on: September 07, 2010, 03:15:23 pm »
reflection in x axis. All the y vlues change sign
so it becomes 4x^2 + 10x + 6 ?

Offline S.M.A.T

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Re: C1 DOUBTS HERE!!!!!
« Reply #243 on: September 08, 2010, 03:53:21 am »
thank you!
i also would like to ask another small question

now when we have an equation y=6-10x-4x^2 and we move the values to the normal way like putting 4x^2 in the beginning then followed by -10x then by -6, does moving the values affect the signs?? if so, how?

I am showing you with an example:
Suppose the question asked to solve the equation 6-10x-4x^2=0
Since before solving we move the values to the normal way like putting 4x^2 in the beginning,so multiply -1 on both side of the '= 'sign i.e (6-10x-4x^2)*-1=0*-1
so the equation become 4x^2+10x-6=0
Not 4x^2 + 10x + 6


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Offline The Golden Girl =D

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Re: C1 DOUBTS HERE!!!!!
« Reply #244 on: September 25, 2010, 05:53:27 pm »
I got a doubt but i know it's a very minute thing but for me it's not cuz idk how much time i've been trying to read books search sites to get to Understand this the ...and the thing that pisses me off is that when my teacher exapleined it he said didn't ya take it in IGCSE Maths ...Man i never EVER saw this thing in my life .

Anyways can some one exaplin to me Completing the Square :'(

http://www.themathpage.com/alg/complete-the-square.htm#complete

cuz i opened this amazing page but i don't get the following :( ;

in  letter c) y did we do + 4 !?!?

e) y did we do 25/4 .... i don't get it ?

f) y did we do 9/4

Plz someone help

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Re: C1 DOUBTS HERE!!!!!
« Reply #245 on: September 25, 2010, 06:01:40 pm »
I got a doubt but i know it's a very minute thing but for me it's not cuz idk how much time i've been trying to read books search sites to get to Understand this the ...and the thing that pisses me off is that when my teacher exapleined it he said didn't ya take it in IGCSE Maths ...Man i never EVER saw this thing in my life .

Anyways can some one exaplin to me Completing the Square :'(

http://www.themathpage.com/alg/complete-the-square.htm#complete

cuz i opened this amazing page but i don't get the following :( ;

in  letter c) y did we do + 4 !?!?

e) y did we do 25/4 .... i don't get it ?

f) y did we do 9/4

Plz someone help



Nvm i get it  :-[
Verily, in the remembrance of Allah do hearts find rest(13:28)

Please, Don't forget to Include GG in your Prayers =D

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Re: C1 DOUBTS HERE!!!!!
« Reply #246 on: September 25, 2010, 06:56:44 pm »
Nvm i get it  :-[

I have a better explanation if you want ?

Offline astarmathsandphysics

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Re: C1 DOUBTS HERE!!!!!
« Reply #248 on: September 26, 2010, 04:25:54 pm »
OMG thx a lot Sir :D

ya sure :)
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Offline astarmathsandphysics

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Re: C1 DOUBTS HERE!!!!!
« Reply #249 on: September 28, 2010, 10:04:48 pm »
The points A (-3,-4) and C (5,4) are the ends of the diagonal of a rhombus ABCD.

a) Find the equation of the diagonal BD.  ( I got x + y = 1)

b) Given that the side BC has gradient 5/3, find the coordinates of B and hence of D.

I don't get part b) at all

Offline Deadly_king

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Re: C1 DOUBTS HERE!!!!!
« Reply #250 on: September 29, 2010, 05:31:02 am »
The points A (-3,-4) and C (5,4) are the ends of the diagonal of a rhombus ABCD.

a) Find the equation of the diagonal BD.  ( I got x + y = 1)

b) Given that the side BC has gradient 5/3, find the coordinates of B and hence of D.

I don't get part b) at all

You got it almost right........except for the part BO.

Anyway let me get this clear for you :

You did obtain equation of line BC ----> 3y = 5x - 13
Now solve simultaneously with x + y = 1 since line BC meets line BD at B.

I'll do it using the substitution method :

From equation 2 --> y = 1 - x
Replace in equation 1 :
3(1 - x) = 5x - 13
3 - 3x = 5x - 13
-8x = -16
x = 2

When x = 2 ----> y = 1 - (2) = -1

Hence coordinates of B (2, -1)

The midpoint of BD = midpoint of AC = (1,0). You calculated it in the first part when finding equation of diagonal BD.

Let coordinates of D be (x,y)
Hence (1,0) = ( (x+2)/2 , (y+-1)/2 )

Therefore (x+2)/2 = 1 ----> x = 0 and (y-1)/2 = 0 ----> y = 1
Coordinates of D (0,1)

Hope it helps :)
« Last Edit: September 29, 2010, 08:31:48 am by Deadly_king »

Offline Ivo

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Re: C1 DOUBTS HERE!!!!!
« Reply #251 on: October 08, 2010, 12:01:55 am »
I'm finding these really hard, please could someone attempt these, many thanks!  Especially explaining why real values of x you state apply.

Q1) The equation of a curve is y=ax^2 - 2bx + c, where a, b and c are constants with a > 0.
a) Find, in terms of a, b and c, the coordinates of the vertex of the curve.
b) Given that the vertex of the curve lies on the line y=x, find an expression for c in terms of a and b.  Show that in this case, whatever the value of b, c >= -1/4a (>= means greater than or equal to).

For this question, I found answers to part a), but am stuck on b).  I know part a) answer is in co-ordinates (b/a , c-(b^2/a)).  I also know some of part b), c is [b(b+1)/a], so I basically don't get why the show that thing is that.

Q2) a) Express 9x^2 +12x + 7 in the form (ax + b)^2 + c where a, b and c are constants whose values are to be found.
b) Find the set of values taken by: 1/(9x^2 +12x + 7)  for real values of x.

For this question, I could get a) with a = 3, b=2, c=3.  But have no idea what "real values of x" mean and please provide reasons why.

Q3) a) Express 9x^2 - 36x + 52 in the form (Ax-B)^2 + C, where A, B and C are integers. Hence, or otherwise, find the set of values taken by 9x^2 - 36x + 52 for real x.  Again I don't know what "real x" means, nor how to work out the answer.

Many thanks in advance, this is too tricky for me ;(
Always willing to help!  8)
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elemis

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Re: C1 DOUBTS HERE!!!!!
« Reply #252 on: October 08, 2010, 04:46:25 am »
Hang on

elemis

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Re: C1 DOUBTS HERE!!!!!
« Reply #253 on: October 08, 2010, 05:22:28 am »
Q2) a) Express 9x^2 +12x + 7 in the form (ax + b)^2 + c where a, b and c are constants whose values are to be found.
b) Find the set of values taken by: 1/(9x^2 +12x + 7)  for real values of x.

For this question, I could get a) with a = 3, b=2, c=3.  But have no idea what "real values of x" mean and please provide reasons why.

COULD SOMEONE PLEASE CHECK THIS EXPLANATION !

Real numbers are any positive or negative numbers which might be whole numbers or numbers like pi or e.

Now you should notice that \frac{1}{(3x+2)^2+3} is the equation of some asymptote. Hence, the value it should be tending to some value, but will never reach it.

Inputing a very large positive or negative number into x we see that it tends towards zero on both cases.

But, I then differentiate the equation to find the maximum points (which is when x = -2/3)

Hence, 0<y\ll\frac{1}{3}

Offline Ivo

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Re: C1 DOUBTS HERE!!!!!
« Reply #254 on: October 08, 2010, 05:22:36 pm »
COULD SOMEONE PLEASE CHECK THIS EXPLANATION !

Real numbers are any positive or negative numbers which might be whole numbers or numbers like pi or e.

Now you should notice that \frac{1}{(3x+2)^2+3} is the equation of some asymptote. Hence, the value it should be tending to some value, but will never reach it.

Inputing a very large positive or negative number into x we see that it tends towards zero on both cases.

But, I then differentiate the equation to find the maximum points (which is when x = -2/3)

Hence, 0<y\ll\frac{1}{3}


OK, thanks for your response.  Your answer/range is right, however I didn't understand why from your last sentence.  Is it because 1/0 doesn't exist, but why 1/3?  What's the general 'rule'?

Also, hardest question is Q1), any ideas, I've thought about it for a long time now.  :(

Thanks for your help so far!
Always willing to help!  8)
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