点 P(2, 3) 和 Q(-1, -3) 在曲线 $y = ax^2 + bx + 2$ 上。列出并求解二元一次方程组，以求出 $a$ 和 $b$ 的值。 答 a = ... b = ...

1. Substitute $P(2,3)$ into the curve equation to form the first equation.

Since $P(2,3)$ lies on $y=ax^2+bx+2$,
$3=a(2)^2+b(2)+2$
$3=4a+2b+2$
$4a+2b=1$

2. Substitute $Q(-1,-3)$ into the curve equation to form the second equation.

Since $Q(-1,-3)$ lies on $y=ax^2+bx+2$,
$-3=a(-1)^2+b(-1)+2$
$-3=a-b+2$
$a-b=-5$

3. Solve the simultaneous equations for $a$ and $b$.

The simultaneous equations are
$4a+2b=1 \quad \text{and} \quad a-b=-5.$
From $a-b=-5$,
$b=a+5.$
Substitute into $4a+2b=1$:
$4a+2(a+5)=1$
$6a+10=1$
$6a=-9$
$a=-\frac{3}{2}.$

4. Find $b$ using the value of $a$.

Substitute $a=-\dfrac{3}{2}$ into $b=a+5$:
$b=-\frac{3}{2}+5$
$b=\frac{7}{2}.$