We have three points (-1,2), (2,-4), (4,2)
The right way to do this problem is just plug in x=-1 into each equation and check that y=2. For any equation that worked for that one try the other two points. I'll do it differently, in a way that doesn't assume we have a menu of choices to choose from.
In general we can work out three equations in three unknowns. But there's a shortcut here because we see two equal y values. As we know the parabola is bilaterally symmetric, symmetric via reflection on its axis. Since we have these equal y values we know the axis is right in the middle, x=(-1+4)/2 = 3/2.
So instead of an arbitrary three variable y=ax²+bx+c we can capture the symmetry and go right to
y = a(x-3/2)² + c
Let's plug in the first two points,
2 = a(-1 - 3/2)² + c = (25/4)a + c
-4 = a(2 - 3/2)² + c = (1/4)a + c
Subtracting,
6 = (24/4)a = 6a
a = 1
-4 = (1/4) + c
c = -17/4
y = (1)(x-3/2)² + -17/4
y = x² - 3x + 9/4 - 17/4
y = x² - 3x - 2
Answer: Choice I