finding quadratic equation, when given 3 points

• Mar 20th 2009, 08:54 PM
kmilne1
finding quadratic equation, when given 3 points
hi. i have the answers to these questions, although i really can't seem to work them out. i know you should sub them into the equation y=ax^2+bx+c but when i have the 3 equations i dont know how to solve them simultaneously. can someone please help?! :)

Find the equation of the quadratic which passes through the points with coordinates:
a
(2, 1), (1, 2), (3, 16)

so far i have:

-1=(-2)^2a - 2b + c
2= a + b + c
-16= 9a + 3b + c

but how do i go further?

• Mar 20th 2009, 09:08 PM
Shyam
Quote:

Originally Posted by kmilne1
hi. i have the answers to these questions, although i really can't seem to work them out. i know you should sub them into the equation y=ax^2+bx+c but when i have the 3 equations i dont know how to solve them simultaneously. can someone please help?! :)

Find the equation of the quadratic which passes through the points with coordinates:
a
(2, 1), (1, 2), (3, 16)

so far i have:

-1=(-2)^2a - 2b + c
2= a + b + c
-16= 9a + 3b + c

but how do i go further?

-1 = 4a - 2b +c .......................(1)
2= a + b + c ....................................(2)
-16= 9a + 3b + c .............................(3)

subtract eqn (2) from (1) that means, (1) - (2) ;
and also subtract eqn (2) from (3) that means, (3) - (2) ; we got,

-3 = 3a -3b .....................(4)
-18 = 8a + 2b ..................(5)

divide eqn (4) by 3 ; and eqn (5) by 2, we got,

-1 = a - b ...............................(6)
-9 = 4a + b .............................(7)

add these eqns (6) and (7)

- 10 = 5a

a = -2

put this value of "a" in eqn (6)
we got, b = -1

now, put these values of "a" and "b" in eqn (2) and find out "c". Finish it.
• Mar 20th 2009, 09:15 PM
kmilne1
Quote:

Originally Posted by Shyam
-1 = 4a - 2b +c .......................(1)
2= a + b + c ....................................(2)
-16= 9a + 3b + c .............................(3)

subtract eqn (2) from (1) that means, (1) - (2) ;
and also subtract eqn (2) from (3) that means, (3) - (2) ; we got,

-3 = 3a -3b .....................(4)
-18 = 8a + 2b ..................(5)

divide eqn (4) by 3 ; and eqn (5) by 2, we got,

-1 = a - b ...............................(6)
-9 = 4a + b .............................(7)

add these eqns (6) and (7)

- 10 = 5a

a = -2

put this value of "a" in eqn (6)
we got, b = -1

now, put these values of "a" and "b" in eqn (2) and find out "c". Finish it.

thanks heaps Shyam :)
but i dont know why you do the bit in red...? (Worried)
why do you need to divide it by the equation?
sorry.

thanks.
• Mar 21st 2009, 02:51 AM
Quote:

Originally Posted by kmilne1
thanks heaps Shyam :)
but i dont know why you do the bit in red...? (Worried)
why do you need to divide the equation?
sorry.

thanks.

Watch the corrected thing in red dividing equation and divide by equation are VERY different things (Wait)
-------
He divided them so that he can cancel out "b"

-3 = 3a -3b .....................(4)
-18 = 8a + 2b ..................(5)

Divide eqn. (4) by 3
-3/3 = 3a/3 - 3b/3

-1 = a - b..............(6) This eqn. is same as (4)

Divide eqn. (5) by 2
-18/2 = 8a/2 + 2b/2

-9 = 4a + b ...........(7) this eqn. is same as (5)