The problem is to analyze the quadratic function $y = ax^2 + bx + c$ given its graph and certain conditions, and to determine the values of $a$, $b$, and $c$ that satisfy the given constraints. Specifically, we are asked to fill in the blanks A through K with the appropriate choices from the given options.

1. Problem Description

The problem is to analyze the quadratic function $y = ax^2 + bx + c$ given its graph and certain conditions, and to determine the values of $a$, $b$, and $c$ that satisfy the given constraints. Specifically, we are asked to fill in the blanks A through K with the appropriate choices from the given options.

2. Solution Steps

(1) Analyzing the graph of $y = ax^2 + bx + c$:

- The parabola opens downward, which means $a < 0$. Thus, A is (9) - "<".

- The graph intersects the x-axis at two points. This is not directly relevant until later.

- When $x = 1$, from the graph $y = a + b + c = 0$. Thus, D is (8) - "=".

- When $x = -1$, from the graph $y = a - b + c = 0$. Thus, E is (8) - "=".

- The vertex appears to be between $x=-1$ and $x=0$, this is not something that can be extracted in the prompt

From $a + b + c = 0$ and $a - b + c = 0$, we can deduce that:

$a + b + c = a - b + c$

$2b = 0$

$b = 0$. Thus, B is (8) - "=".

Since $a+c=0$ and $a \ne 0$, then $c = -a$ and since $a < 0$, $c > 0$. Thus, C is (7) - ">".

$4a + 2b + c = 4a + 0 + c = 4a + c = 4a - a = 3a$. Since $a<0$, $3a < 0$, so $4a+2b+c < 0$, F is (9) - "<". This is not needed.

The discriminant $b^2 - 4ac = 0^2 - 4a(-a) = 4a^2 > 0$ since $a \ne 0$. Thus G is (7) - ">". This is also not needed.

(2) Minimizing $a^2 - 8b - 8c$ with $a < 0$, $b = 0$, and $c = -a$.

Substituting $b = 0$ and $c = -a$ into the expression, we have:

$a^2 - 8(0) - 8(-a) = a^2 + 8a$.

To minimize $f(a) = a^2 + 8a$, we can find the vertex of the parabola.

The vertex is at $a = -\frac{8}{2(1)} = -4$. Since $a<0$, $a=-4$ is valid.

Thus, H is (6) - "-4".

When $a = -4$ and $b = 0$, then $y = -4x^2 + 0x + c$. Also $c = -a$, so $c = 4$.

So $y = -4x^2 + 4$.

We are told that $y=ax^2+bx+c$ can be written as $y = ax^2+bx-b+I$, and $a=-4, b=0, c=4$, so $y = -4x^2 + 0x + 4$, then $c=-b+I$ means $4=-0+I$, hence $I=4$.

Thus I is (4) - "4".

We have $a = -4$ and $b = 0$.

Since we need to maintain the condition $a+b+c = 0$ we have that $b$ can be any value and we need $c=-a-b$ which keeps $a+b+c = a+b-a-b=0$.

The problem statement says we need (i) a<0 and (ii) $a+b+c=0$. Then $c = -a-b$ and we want to minimize $a^2-8b-8c=a^2-8b-8(-a-b) = a^2-8b+8a+8b = a^2+8a$. So a needs to be the minimum, $a=-4$. The range of $b$ that satisfies these equations is not restricted, it appears, by previous results, but we know that $a-b+c=a-b-a-b = -2b$ which means that $-2b=0$ for $a-b+c=0$, $b=0$, thus $-2b<0$ so b is unrestricted. Now we are told that $y = ax^2+bx+c$ where $a<0$, $b=0, c>0$, $a=-4, b=0, c=4$ from above. Then y can be expressed as $y = -4x^2+bx-b+4$ when $y = ax^2 + bx + c = -4x^2+0x+4$, $-b+4=4$ and $b=0$. Since the parabola intersects x-axis at -1 and 1, it should be such that $a<0$, and parabola $y = ax^2+bx+c$ satisfies $y<0$, $b \ge 0$ and $b$ can not be greater than 1,

$0= -4+b$, $b = 4$, so

We need $b$ such that $a<0,$b=$a/2$,

Since $a<0$, $0<b<1$.

Let's look for other constraints from a-b+c=0 and a+b+c=

0. a-b+c=0 and a+b+c=0 imply $a+c=b$ and $a+c=-b$, $b = -b$ means $b=0$.

b must be 0, but we also have to consider c must be greater than zero,

$a<0 , b=0, c = -a>0$. If $b=0$, than

$0<b<1$, so from these considerations we have, J is 0 and K is

1. Thus J is (0) and K is (1).

3. Final Answer

A: (9)

B: (8)

C: (7)

D: (8)

E: (8)

F: (9)

G: (7)

H: (6)

I: (4)

J: (0)

K: (1)