The problem is about a quadratic function $y = ax^2 + bx + c$ and its graph. We need to determine the signs of $a$, $b$, and $c$ based on the graph, and then use the given equations to find the values of $a$, $c$ in terms of $b$. Finally, we need to find the range of possible values for $b$ when $a^2 - 8b - 8c$ is minimized.

1. Problem Description

The problem is about a quadratic function $y = ax^2 + bx + c$ and its graph. We need to determine the signs of $a$, $b$, and $c$ based on the graph, and then use the given equations to find the values of $a$, $c$ in terms of $b$. Finally, we need to find the range of possible values for $b$ when $a^2 - 8b - 8c$ is minimized.

2. Solution Steps

(1)

(i) Since the parabola opens downwards, $a < 0$. The graph intersects the y-axis at a positive value, so $c > 0$.

Since the axis of symmetry is on the positive side of y-axis, $-b/2a > 0$. Since $a < 0$, this implies $b > 0$.

So, $A$ is (9), $B$ is (7), and $C$ is (7).

(ii) The problem says $a+b+c = 0$

Thus, $D$ is (0)

(iii) The problem says $a-b+c=0$

Thus, $E$ is (0)

(iv) The image does not give enough information to solve this equation.

(v) The image does not give enough information to solve this equation.

(2) From (i) and (ii), we have:

$a+b+c = 0$

$a-b+c = 0$

Subtracting the second equation from the first, we get $2b = 0$, so $b = 0$. However, we found that $b > 0$ in part (1), and therefore, we cannot use the equation $a+b+c = 0$ directly.

Since the problem mentions (i), we use $a < 0$, $b > 0$, and $c > 0$. From (ii) and (iii), we have $a+c = -b$ and $a+c=b$. Thus $b = -b$, so $2b=0$, and $b=0$.

This means our initial assessment was wrong.

We have $a+b+c = 0$ and $a-b+c=0$. Then $2a+2c=0$ or $a+c = 0$.

Then $a = -c$.

$a^2 - 8b - 8c = (-c)^2 - 8b - 8c = c^2 - 8b - 8c$

Since $a+b+c = 0$, $a=-b-c$.

Since $a-b+c=0$, $a=b-c$.

Thus $-b-c = b-c$, so $-b = b$, which implies $2b = 0$ or $b=0$.

If $b = 0$, we have $a+c = 0$ or $a = -c$. Then $a^2 - 8b - 8c = a^2 - 8c = (-c)^2 - 8c = c^2 - 8c$.

The value is minimized when $2c - 8 = 0$, so $c = 4$.

Then $a = -4$.

However, we need to express $a$ in terms of $b$. If $a = b-c$, then $a=-4$, and $b=0$, so $c = 4$.

Since $a=-c$, and $a<0$ and $c>0$, we consider the function $f(a,b,c) = a^2 - 8b - 8c$.

Since $a+b+c = 0$, we have $c = -a - b$.

Substituting for $c$, we have $a^2 - 8b - 8(-a-b) = a^2 - 8b + 8a + 8b = a^2 + 8a$.

The minimum of $a^2 + 8a$ occurs when $2a+8=0$, so $a=-4$.

Then $H$ is (6).

$y = -4x^2 + bx + c = -4x^2 + bx - b + I$.

Since $a+b+c = 0$, we have $-4+b+c=0$, so $c = 4-b$.

Therefore, $y = -4x^2 + bx + 4-b$.

Thus, $I$ is (4).

Also $b>0$. Since $a+b+c=0$ and $a-b+c=0$, $a+c = -b$, and $a+c=b$, so $2b = 0$. This is not correct.

The vertex must be $0 < x < 1$, so $0 < -b/(2a) < 1$. Since $a<0$, we have $2a < -b < 0$, or $0 < b < -2a$.

Since $a=-4$, we have $0 < b < 8$.

Since $a = -4$, $a < 0$. $a+b+c=0$, $-4+b+c=0$, so $c=4-b$. Since $c>0$, $4-b>0$, so $b<4$.

Therefore $0<b<4$.

$J$ is (0), and $K$ is (4).

3. Final Answer

A: (9)

B: (7)

C: (7)

D: (0)

E: (0)

F: Don't Know

G: Don't Know

H: (6)

I: (4)

J: (0)

K: (4)