点 A(0,2) 和 B(1,0) 在函数图象 $y = x^2 + ax + b$ 上。求 a 和 b 的值。 答案 a = ... b = ... C 是图象上的另一个点，使得直线 AB 的斜率是 AC 斜率的两倍。求点 C 的坐标。 答案 (... , ...)

1. Use point A to find $b$.

Since $A(0,2)$ lies on $y=x^2+ax+b$, substitute $x=0$ and $y=2$:

$2=0^2+a(0)+b$

$b=2$

2. Use point B and the value of $b$ to find $a$.

Since $B(1,0)$ lies on the graph, substitute $x=1$, $y=0$ and $b=2$:

$0=1^2+a(1)+2$

$0=3+a$

$a=-3$

1. Find the gradient of line $AB$ and hence the gradient of line $AC$.

Using $A(0,2)$ and $B(1,0)$,
$ext{gradient of }AB=\frac{0-2}{1-0}=-2.$
Given that the gradient of $AB$ is twice the gradient of $AC$,
$-2=2(\text{gradient of }AC)$
$\text{gradient of }AC=-1.$

2. Form the equation of line $AC$ through $A(0,2)$ with gradient $-1$.

Since line $AC$ passes through $A(0,2)$ and has gradient $-1$,
$y-2=-1(x-0)$
$y=2-x.$

3. Find the other intersection of line $AC$ with the graph $y=x^2-3x+2$.

From part (a), the graph is
$y=x^2-3x+2.$
At point $C$, it also lies on $y=2-x$, so
$x^2-3x+2=2-x$
$x^2-2x=0$
$x(x-2)=0.$
Thus $x=0$ or $x=2$.

$x=0$ gives point $A$, so the other point is $x=2$.
$y=2-2=0.$