system of linear equations curve fitting?

Question

Find the equation y = ax^2 +bx +c whose graph contains the points (1, -1) (3, -1) (-2, 14)

you have to solve using the system of linear equation stuff, so I did:

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

and got the system:
a+b+c = -1
9a+3b+c = -1
4a-2b+c = 14

so I went through and solved:
first with lines 1 and 2:
-9a-9b-9c = 9
9a +3b+c=-1 and got -6b-8c=8 . . . plugged back into system

a+b+c = -1
-6b-8c=8
4a-2b+c = 14 then took lines 1 and 3 did the same:

-4a-4b-4c = 4
4a-2b+c=14 and got -6b-3c=18 plugged back into system:

a+b+c = -1
-6b-8c=8
-6b-3c=18 then took lines 2 and 3 did the same:

and got c = -3, pluged that back into a different equation and got b = 13/3 and plugged both into first equation and got a = -1/3

now did I do this correctly???

please let me know!! thanks!!

Answer 1

In your system in which -6b - 8c = 8 and -6b - 3c = 18, multiply the second by -1 to get

-6b -8c = 8
6b + 3c = -18

adding these gives -5c = -10, so c = 2. You have c = 3

Your approach is correct, however--check the arithmetic.

Answer 2

Whenever you do a problem like this, you can see for yourself if it is correct by putting the popints in and seeing if the equation is OK. You have:

(-1/3)x^2 + (13/3) x - 3 = y

Try (1, -1) : -1/3 + 13/3 - 3 = 1 Checks
Try (3, -1) -9/3 + 13 - 3 = -1 Oops this adds to 7 not -1
Try (-2, 14) -4/3 - 26/3 - 3 = 14 Oops again, it adds to -13

Your work looks fine up until "and got c= -3"

You have:

-6b-8c=8
-6b-3c=18

Subtract these: , you get:

-5c = -10 so c = 2, then b = -4, a = 1

Your technique is fine but be careful on jumping steps without writing it all out. Careful documentation of the steps makes it much easier to find you errors.