y = ax² + bx + c

The fact that it has x-intercepts of 0 and 1 tells us that the parabola
goes through the points (0,0) and (1,0)

So we substitute (0,0)

0 = a(0)^2 + b(0) + c
0 = 0 + 0 + c 
0 = c

So we replace c by 0

y = ax² + bx + c
y = ax² + bx + 0
y = ax² + bx

And we substitute (1,0)

y = ax² + bx
0 = a(1)² + b(1)
0 = a(1) + b
0 = a + b
a + b = 0
    b = -a

So we substitute -a for b

y = ax↨ _ bx
y = ax² - ax 

And we substitute (2,-2)

y = ax² - ax
-2 = a(2)² - a(2)
-2 = a(4) - 2a
-2 = 4a - 2a
-2 = 2a
-1 = a

Substitute -1 for a

y = (-1)x²-(-1)x
y = -1x²+1x
y = -x²+x

Here's the graph.  Notice that it passes through (0,0), (1,0), and (2,-2)





Edwin