The problem involves two quadratic functions $f(x) = -2x^2$ and $g(x) = x^2 + ax + b$. We are given two conditions: (i) $g(x)$ has a minimum value at $x=3$, and (ii) $g(4) = f(4)$. We need to find values for $a$, $b$, the minimum value of $g(x)$, the solution to $f(x)=g(x)$ other than $x=4$, and the maximum value of $f(x)-g(x)$ on the interval $[-H, 4]$.

1. Problem Description

The problem involves two quadratic functions $f(x) = -2x^2$ and $g(x) = x^2 + ax + b$. We are given two conditions: (i) $g(x)$ has a minimum value at $x=3$, and (ii) $g(4) = f(4)$. We need to find values for $a$, $b$, the minimum value of $g(x)$, the solution to $f(x)=g(x)$ other than $x=4$, and the maximum value of $f(x)-g(x)$ on the interval $[-H, 4]$.

2. Solution Steps

(1) Condition (i) implies that the vertex of the parabola $g(x) = x^2 + ax + b$ occurs at $x=3$. The $x$-coordinate of the vertex of a parabola $x^2 + ax + b$ is given by $x = -\frac{a}{2}$. Thus, $-\frac{a}{2} = 3$, so $a = -6$.

$A = 6$

Condition (ii) states that $g(4) = f(4)$. We have $f(4) = -2(4^2) = -2(16) = -32$. Also, $g(4) = (4)^2 + a(4) + b = 16 + 4a + b$. Substituting $a = -6$, we get $g(4) = 16 + 4(-6) + b = 16 - 24 + b = -8 + b$.

Since $g(4) = f(4)$, we have $-8 + b = -32$, so $b = -32 + 8 = -24$.

Thus, $BC = 24$.

Now, we can write $g(x) = x^2 - 6x - 24$. To find the minimum value, we complete the square:

$g(x) = (x^2 - 6x + 9) - 9 - 24 = (x-3)^2 - 33$. The minimum value of $g(x)$ is $-33$.

$DE = 33$

(2) To find the values of $x$ such that $f(x) = g(x)$, we set $-2x^2 = x^2 - 6x - 24$.

This simplifies to $0 = 3x^2 - 6x - 24$. Dividing by 3, we get $x^2 - 2x - 8 = 0$.

Comparing this to $x^2 - Fx - G = 0$, we have $F=2$ and $G=8$.

Factoring the quadratic $x^2 - 2x - 8 = 0$, we get $(x-4)(x+2) = 0$. Thus, the solutions are $x = 4$ and $x = -2$.

Since we want the solution other than $x=4$, we have $x = -2$.

$H = 2$

(3) We want to find the maximum value of $f(x) - g(x)$ on the interval $[-2, 4]$.

$f(x) - g(x) = -2x^2 - (x^2 - 6x - 24) = -3x^2 + 6x + 24$.

Let $h(x) = -3x^2 + 6x + 24$. To find the vertex, we use $x = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1$.

$h(1) = -3(1)^2 + 6(1) + 24 = -3 + 6 + 24 = 27$.

Since the parabola opens downward, the maximum value occurs at the vertex. Since the vertex $x=1$ lies within the interval $[-2, 4]$, the maximum value is $27$ at $x=1$.

So, $I = 1$.

The maximum value is $27$, so $JK = 27$.

3. Final Answer

A = 6

BC = 24

DE = 33

F = 2

G = 8

H = 2

I = 1

JK = 27