I take it g(x) is given by this description:
a parabola that opens down and passes through 0 comma 3, 3 comma 12, and 5 comma 8
If we could see the figure we'd known if (3,12) was the max they're talking about. We'll have to check.
Opens down is a CDN, concave down negative, so a negative coefficient on the squared term. Through (0,3), (3,12), (5,8)
g(x) = ax² + bx + c
3 = g(0) = c
g(x) = ax² + bx + 3
12=g(3) = 9a + 3b + 3
9 = 9a+3b
3 = 3a + b
8=g(5) = 25a + 5b +3
5 = 25a + 5b
1 = 5a + b
Subtracting the two equations '3=' and '1=' we get
2 = -2a
a = -1
b = 1 - 5a = 6
g(x) = -x² + 6x + 3 = -(x² - 6x) + 3 = -(x² - 6x + 9) + 3 + 9 = 12-(x-3)²
and we see the max at x=3, y=12. It would have been easier to show the graph or say that.
f(x) = x² - 6x + 3 = (x - 3)² + 3 - 9 = (x-3)² - 6
So this mas a minimum (the squared term can't get any less than zero) at x=3, y=-6
Distance between them is 12 - -6 = 18
Answer: 18
