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# finding the equation - c1 quick question Watch

1. hey guys this is a question for c1 edexcel.

My questions

Q1) I know we are trying to find the equation and since its a quadratic it will be in the form (x-p) (x-q) but i dnt get why there is an 'a' in front of it?

Q2) secondly is it totally nessary to write out y=a(x-p)(x-q) ?

Q3)lastly can someone explain every step in"Using (2,-1) ----> -1=a(1)(-1) ----> a=1"

Thanks
2. (Original post by jayseanfan)
hey guys this is a question for c1 edexcel.

My questions

Q1) I know we are trying to find the equation and since its a quadratic it will be in the form (x-p) (x-q) but i dnt get why there is an 'a' in front of it?

Q2) secondly is it totally nessary to write out y=a(x-p)(x-q) ?

Q3)lastly can someone explain every step in"Using (2,-1) ----> -1=a(1)(-1) ----> a=1"

Thanks
1) you need the a becos you need to know the coefficient to know if its stretched or what not. 2) yes its necessary
3) you sub your point (2,-1) in your equation (which u should know should be y=a(x-1)(x-3) from your graph ). do this to find your a
3. can somone help with this , i didnt understand the above explanation
4. (Original post by jayseanfan)
can somone help with this , i didnt understand the above explanation
The a is in front because it tells you if the graph has been stretched or squashed undergone a transformation
5. (Original post by jayseanfan)
can somone help with this , i didnt understand the above explanation
1) if you didn't have the 'a', it would not allow for the x^2 term to have a coefficient other than 1 (ie - a polynomial with 2x^2 as a leading term is still quadratic.)

2) that is one way you could do it. it is probably the easiest way in the context of this question.

3) (2,-1) means that when x=2, f(x) = -1. So you substitute x=2 and y=-1 into a(x-1)(x-3) = y, then you have an equation where 'a' is the only unknown, allowing you to solve for 'a'.

Hope this helps.
6. (Original post by jayseanfan)
can somone help with this , i didnt understand the above explanation
1. You need 'a' in front of it because for instance,
y=2(x-p)(x-q) and y=5(x-p)(x-q) would look different, right?
but they both pass (p,0) and (q,0)

2. Yes

3. Put (2, -1) into y=a(x-1)(x-3)
-1 = a (2-1)(2-3)
-1 = a*1*-1

7. (Original post by gtmad)
1) if you didn't have the 'a', it would not allow for the x^2 term to have a coefficient other than 1 (ie - a polynomial with 2x^2 as a leading term is still quadratic.)

2) that is one way you could do it. it is probably the easiest way in the context of this question.

3) (2,-1) means that when x=2, f(x) = -1. So you substitute x=2 and y=-1 into a(x-1)(x-3) = y, then you have an equation where 'a' is the only unknown, allowing you to solve for 'a'.

Hope this helps.
Thank you guys for your replies. I get it now. Sorry this might be a stupid question but do you have to use 'a' 'p' and 'q' or can you use any letters?
8. (Original post by jayseanfan)
Thank you guys for your replies. I get it now. Sorry this might be a stupid question but do you have to use 'a' 'p' and 'q' or can you use any letters?
you can use any letters but many people use a p q I think

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