C1 and C2 DOUBTS HERE!!!!!

[Deadly_king]

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!

Sorry for late reply guys 

Yeah....Ari is right there

Hmm.........from part(i) of the question, you can find the turning point to be (-2/3 , 3)
So you should replace the value of x in the equation to get the range of value of y. Since (3x + 2)² is a perfect square......y will in no way be less than zero.

[Ivo]

Thanks for your help guys, this question is still bothering me though.  Whoever can solve this is a genius!  I can't believe this question was from C1.

The equation of a curve is , where a, b and c are constants with .

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 , find an expression for c in terms of a and b. Show that in this case, whatever the value of b, .

I've partially answered the question (I've done parts a) and a bit of b), but don't know how to show that the statement in b) is correct).

Here are my answers so far:

a)

b) , but I don't know how to 'show that...'.

Many many thanks in advance!

[elemis]

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'?

Since it is some form of an asymptote the Y value will go closer and closer to zero but NEVER reach zero.

Remember I found the max point to be -2/3 ? That was the x coordinate.

Inputting -2/3 into the original equation gives a Y VALUE of 1/3

Since this is the max point the curve's maximum value that it can attain is 1/3

[**RoRo**]

OKAY, here's a question that has been driving me nuts since last week:

Solve the simultaneous equations:
2x + 2y = 7
x² - 4y² = 8

The answers according to the textbook are: x=3, y=1/2 OR x=6 1/3, y= -2 5/6

Thanks a lot in advance!

[Dibss]

Solve the simultaneous equations:
2x + 2y = 7
x² - 4y² = 8

First eq rearranged.
x= (7-2y)/2
Substitute the x in the second eq.
[(7-2y)²]/2² - 4y² = 8
(49-28y+4y²)/4 -4y²= 8
49 - 28y + 4y² -16y² = 32
Now just solve like a normal quadratic.
Did this in a hurry, will check up the answer and get back to you in a few (:

[Dibss]

First eq rearranged.
x= (7-2y)/2
Substitute the x in the second eq.
[(7-2y)²]/2² - 4y² = 8
(49-28y+4y²)/4 -4y²= 8
49 - 28y + 4y² -16y² = 32
Now just solve like a normal quadratic.
Did this in a hurry, will check up the answer and get back to you in a few (:

Okay so you get the eq
12y² + 28y -17 = 0
Solve it
y = 0.5 or y = 2.83
Substitute values into the first eq and get x values.

Makes sense or you have any questions? (:

[Dibss]

Heyy Ivo, let me try (:

[Dibss]

Okay being the blondie that I am, solved the first two parts in a minute then got stuck on the proving for eternity. Decided to post up at least the parts I managed and read that you already have the same answers xP
Sorry, can't be of help - I'm as lost as you :/

[Ivo]

Okay being the blondie that I am, solved the first two parts in a minute then got stuck on the proving for eternity. Decided to post up at least the parts I managed and read that you already have the same answers xP
Sorry, can't be of help - I'm as lost as you :/

Well at least I have company!    Anyway, thanks for your attempt.  Now I know I'm not the only one!

[Freaked12]

Yo tell me if this solution is okay



(i) y = ax^2 - 2bx + c, we differentiate to get:
dy/dx = 2ax - 2b.

We know dy/dx=0 at the vertex, we get the coordinates you asked for
(ii) Now given that the vertex is on y=x, this tells as that:
b/a = c - (b^2/a) => c= (b(b+1))/a

This equation relates 'c' with 'a' and 'b'.. Also the factor of 4 and => sign might make you think of a discriminant, why?

The equation for 'c' is in fact a quadratic in 'b':
b^2 + b - ac = 0.

 We require D => 0 for real roots to exist, and since we are dealing with a real cartesian system this has to be the only case, therefore:

(1)^2 - 4*(1)*(-ac) >= 0 => 1 + 4ac >=0
and so we get c >= -1/(4a), since we know a>0/

[Dibss]

We require D >= 0 for real roots to exist, and since we are dealing with a real cartesian system this has to be the only case, therefore:

(1)^2 - 4*(1)*(-ac) >= 0 => 1 + 4ac >=0

Can't believe I didn't figure that out :/
+Rep (:

[Freaked12]

Even i didnt know until i copied this question down to my tutor  so he could explain it to me

[Ivo]

Perfect answer, and your method would make sense in context with the question.  This question came out of a chapter in the textbook dealing with quadratics - more specifically completing the square and about discriminant .  So this question ties the two into one quite nicely!

Thank you for your (and that of your tutor)'s efforts - greatly appreciated and now I understand why!  I will + rep you, definitely!!!  

[Deadly_king]

Yo tell me if this solution is okay



(i) y = ax^2 - 2bx + c, we differentiate to get:
dy/dx = 2ax - 2b.

We know dy/dx=0 at the vertex, we get the coordinates you asked for
(ii) Now given that the vertex is on y=x, this tells as that:
b/a = c - (b^2/a) => c= (b(b+1))/a

This equation relates 'c' with 'a' and 'b'.. Also the factor of 4 and => sign might make you think of a discriminant, why?

The equation for 'c' is in fact a quadratic in 'b':
b^2 + b - ac = 0.

 We require D => 0 for real roots to exist, and since we are dealing with a real cartesian system this has to be the only case, therefore:

(1)^2 - 4*(1)*(-ac) >= 0 => 1 + 4ac >=0
and so we get c >= -1/(4a), since we know a>0/

Good job pal

I got stuck at the same place as Ivo........I thought about the discriminant part but could not quite link it  

+ rep btu i need to spread the love first 

[astarmathsandphysics]

I missed that one. Congrats