What is the equation in standard form of a parabola that contains the following points
(-2,-16) (0,4) (4-28)

Equation of a parabola is given by
y = ax^2 + bx + c
From the given points
-16 = a(-2)^2 + b(-2) + c = 4a - 2b + c . . . (1)
4 = a(0)^2 + b(0) + c = c . . . (2)
-28 = a(4)^2 + b(4) + c = 16a + 4b + c . . . (3)

Putting (2) into (1) and (2) gives:
4a - 2b + 4 = -16
4a - 2b = -20 . . . (4)
16a + 4b + 4 = -28
16a + 4b = -32 . . . (5)
(4) x 4 => 16a - 8b = -80 . . . (6)
(5) - (6) => 12b = 48
b = 48/12 = 4

From (4), 4a - 2(4) = -20
4a = -20 + 8 = -12
a = -12/4 = -3

Therefore, the required polynomial is
y = -3x^2 + 4x + 4

What’s the question?