Question

PLEASE HELP!!!!! I am giving 15 points

Answer 1

Answer:

Step-by-step explanation:

I used 3 of the coordinates to solve for the a, b, and c that we need in

$y=ax^2+bx+c$

We have 3 unknowns so we need 3 equations.  The coordinates I used are from your table:

(-2, 12), (-1, 6), (0, 2).  Start with (0, 2).  Sub in a 2 for y and a 0 for x in the standard form of the quadratic:

$2=a(0)^2+b(0)+c$ which simplifies very nicely to

c = 2.  We will use that value now when we do this again using (-2, 12):

$12=a(-2)^2+b(-2)+2$  and

12 = 4a - 2b + 2 and

4a - 2b = 10 (1)

Do it again with the third coordinate (-1, 6):

$6=a(-1)^2+b(-1)+2$ and

6 = 1a - b + 2 and

a - b = 4 (2)

Now we have a system of equations, (1) and (2) that we will combine and solve for a:

 4a  -  2b  =  10

   a   -   b   =   4

Multiply the second equation by -2 to get a new system:

 4a  -  2b  =  10

-2a  +  2b = -8

Add to get

2a = 2 so

a = 1

Now sub in 1 for a in (2):

1 - b = 4 and

-b = 3 so

b = -3

Now we have all that we need to write the equation:

$y=x^2-3x+2$