图中显示了曲线 y = x^2 + px + q。点 A(0, −5) 和 B(2, −9) 位于图像上。B(2, −9) 是最低点。

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完整解法

1. Use point A to find $q$.

Since $A(0,-5)$ lies on the curve,
$-5=0^2+p(0)+q$
$q=-5$

2. Use point B to form an equation involving $p$.

Since $B(2,-9)$ lies on the curve,
$-9=2^2+p(2)+q$
$-9=4+2p+q$

3. Substitute $q=-5$ and solve for $p$.

Substitute $q=-5$ :
$-9=4+2p-5$
$-9=2p-1$
$2p=-8$
$p=-4$

答案

$p=-4$
$q=-5$

1. Find the gradient of line $BC$.

For $B(2,-9)$ and $C(7,16)$ ,
$m_{BC}=\frac{16-(-9)}{7-2}=\frac{25}{5}=5.$

2. Use the same gradient and point $A$ to find the required equation.

The required line is parallel to $BC$ , so its gradient is also $5$ .

It passes through $A(0,-5)$ , so the $y$ -intercept is $-5$ .

Hence,
$y=5x-5.$

答案

$y=5x-5$

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