#1

Can somebody help me with my AS level maths revision please?

i want to know how to complete the square to find the vertex of a parabola.

let's say it is given in the form y=ax^2+bx+c

i need to know an easy, practical method to complete the square, from there i think i should be able to find the vertex

thanks

i want to know how to complete the square to find the vertex of a parabola.

let's say it is given in the form y=ax^2+bx+c

i need to know an easy, practical method to complete the square, from there i think i should be able to find the vertex

thanks

#2

isn't it something like -b/2a to find the vertex?

#3

y=ax^2+bx+c

y=(SQRTx+1/2b)+ whatever it takes to make (1/2b)^2=c

I'm pretty sure it's that anyway

Please note the edit

y=(SQRTx+1/2b)+ whatever it takes to make (1/2b)^2=c

I'm pretty sure it's that anyway

Please note the edit

#4

umm that sounds like your talking about the quadratic function, the quadratic function will give you the x intercepts (if any)i need to find the vertex (the very minimum point on a parabola curve.

thanks anyway

thanks anyway

#5

y=ax^2+bx+c

y=(SQRTx+1/2b)+ whatever it takes to make b^2=c

I'm pretty sure it's that anyway

ok that sounds like it, could you give me an example so i know where i'm going, for instance if y=x^2+4x+10 what would be the complete square form?

thanks

*Last edited by 3nemy at Nov 23, 2007,*

#6

ok that sounds like it, could you give me an example so i know where i'm going, for instance if y=2x^2+4x+10 what would be the complete square form?

thanks

y=(SQRT2x+2)+6

I think anyway, I ought to be revising this, I have my C1 and C2 retakes in Jan -_-

#7

Can you use derivative? It's the only thing I can think just now...

#8

Ah, completing the square.

Take the quadratic function f(x) = ax^2 + bx + c = 0

Take the negative half of b and place it within brackets (x - (b/2))^2.

Now, this isn't equal to what we started with, the only difference, is the constant.

So, work out the constant. (b/2)^2. A subtract this from your equation.

(x - (b/2))^2 - (b/2)^2. What you now need to do, now you've got rid of the constant. Is add the original constant to the equation, so we're left with what we started with in a different form.

(x - (b/2))^2 - (b/2)^2 + c = 0.

:-) and there you go.

It's a little different when a > 1 but this should work for when x = 1. If you wish to learn more, add me. matthew_haworth@hotmail.com, I take A2 mathematics and further mathematics, so i like to help.

Take the quadratic function f(x) = ax^2 + bx + c = 0

Take the negative half of b and place it within brackets (x - (b/2))^2.

Now, this isn't equal to what we started with, the only difference, is the constant.

So, work out the constant. (b/2)^2. A subtract this from your equation.

(x - (b/2))^2 - (b/2)^2. What you now need to do, now you've got rid of the constant. Is add the original constant to the equation, so we're left with what we started with in a different form.

(x - (b/2))^2 - (b/2)^2 + c = 0.

:-) and there you go.

It's a little different when a > 1 but this should work for when x = 1. If you wish to learn more, add me. matthew_haworth@hotmail.com, I take A2 mathematics and further mathematics, so i like to help.

*Last edited by matth05 at Nov 23, 2007,*

#9

this might help, also

#10

ok so am i right in saying the complete square form of x^2+4x+10=0 is (x-2)^2+6?

thanks for all your help

thanks for all your help

#11

I dont remember any of that **** you guys are talking about and I just learned how to do this a few weeks ago.

With your example of y=2x^2+4x+10...

You first take that 4 of the x, and divide it by 2, then square it. So basically in this case, you will still get 4.

So with that new number 4, write out the equation as y=2x^2+4x

So the first three are 2x^2+4x+4. When reduced we get (2x+2)(x+2)...................

Actually dude, you cant do it with that 2x^2, unless Im doing something else here. To complete the square, you have to have just x^2.

EDIT: damn, im a little slow. and btw guys, Ive never heard of that SQRT ****

With your example of y=2x^2+4x+10...

You first take that 4 of the x, and divide it by 2, then square it. So basically in this case, you will still get 4.

So with that new number 4, write out the equation as y=2x^2+4x

**+4-4**+10. What this enables you to do is reduce the first 3 numbers in that equation.So the first three are 2x^2+4x+4. When reduced we get (2x+2)(x+2)...................

Actually dude, you cant do it with that 2x^2, unless Im doing something else here. To complete the square, you have to have just x^2.

EDIT: damn, im a little slow. and btw guys, Ive never heard of that SQRT ****

#12

oh forget that 2 in front of the x^2 in my original question, that shouldn't be there it is either +/-

EDIT: can i just ask what determines the +/- value of the b in the brackets or is it always -?

thanks

EDIT: can i just ask what determines the +/- value of the b in the brackets or is it always -?

thanks

#13

oh forget that 2 in front of the x^2 in my original question, that shouldn't be there it is either +/-

EDIT: can i just ask what determines the +/- value of the b in the brackets or is it always -?

thanks

I've just come to realise how bad my explanation is, if you want an indepth explanation of any maths concept, I'll genuinely be happy to help :-).