How to find the quadratic model for (-2,-14) (0,2) (4,-38)

Question

anonymous · Answer

Write
$$
f(x)= a x^2+b x+c
$$

anonymous · Answer

You have to find a , b c
Solve
$$
f(-2)=-14\
f[0]=2\
f(4) =-38
$$
You have to solve 3 equations in three unknown. The 2nd equation
gives c=2. So it is only two equations in two of them. Solve them and find a and b

triciaal · Answer

when i solved i got a = -3; b=-2 and c = 2 which would make 
f(x) -3x^2 -2x +2

triciaal · Answer

eqn 3
-38 = a(4)^2 + b(4) + 2
-10 = 4a + b
eqn 1
-14 = a(-2)^2 + b(-2) +2
-16 = 4a -2b
-8 = 2a - b 

add the 2 eqn  -18 = 6a  so a = -3
find b using eqn 1  -8 =2(-3) -b   -8 + 6 = -2

a= -3; b = -2; c =2

anonymous · Answer

By the way b=2 and this is the final answer
$$f(x)=-3 x^2+2 x+2$$
Here is the graph of f and the points