Answer:
Step-by-step explanation:
As the slope between (1,0) and (2,2) is 2 and the slope between (2,2) and (4,4) is 1, we can tell that the parabola must be downward opening and have a vertex in the first quadrant
Possibly best to use the vertex form to make 3 equations with 3 unknowns
y = a(x - h)² + k
0 = a(1 - h)² + k (i)
2 = a(2- h)² + k (ii)
4 = a(4 - h)² + k (iii)
subtract i from ii
2 = a((2 - h)² - (1 - h)²)
2 = a(4 - 4h + h² - (1 - 2h + h²))
2 = a(3 - 2h)
a = 2/(3 - 2h)
subtract i from iii
4 = a((4 - h)² - (1 - h)²)
4 = a(16 - 8h + h² - (1 - 2h + h²))
4 = a(15 - 6h)
substitute for a
4 = (2/(3 - 2h))(15 - 6h)
4(3 - 2h) = 2(15 - 6h)
12 - 8h = 30 - 12h
4h = 18
h = 9/2
a = 2/(3 - 2(9/2))
a = - 1/3
0 = -1/3(1 - 9/2)² + k
k = 49/12
y = (-1/3)(x - 9/2)² + 49/12
expand to get y = ax² + bx + c form
y = (-1/3)(x² - 9x + 81/4) + 49/12
y = (-1/3)x² + 3x - 27/4) + 49/12
y = (-1/3)x² + 3x - 8/3
