点 (4,20) 和 (10,-4) 在由方程 $y = ax^2 + bx - 4$ 确定的曲线上。 用代数方法确定 $a$ 和 $b$ 的值。 答案 a = ... , b = ...

1. Substitute $(4,20)$ into the curve equation to form the first equation.

Since $(4,20)$ satisfies $y=ax^2+bx-4$,
$20=a(4)^2+b(4)-4$
$20=16a+4b-4$
$16a+4b=24$
$4a+b=6 \quad \text{--- (1)}$

2. Substitute $(10,-4)$ into the curve equation to form the second equation.

Since $(10,-4)$ satisfies $y=ax^2+bx-4$,
$-4=a(10)^2+b(10)-4$
$-4=100a+10b-4$
$100a+10b=0$
$10a+b=0 \quad \text{--- (2)}$

3. Eliminate $b$ to find $a$.

Subtract (1) from (2):
$(10a+b)-(4a+b)=0-6$
$6a=-6$
$a=-1$

4. Substitute $a$ back into one equation to find $b$.

Using (2),
$10a+b=0$
$10(-1)+b=0$
$b=10$