Leanne 说直线 AB 的方程是 $2y = -4x + 1$，直线 PQ 的方程是 $4y = -9x + 2$。 她说得对吗？解释你的答案。 图中显示了 $y = x^2 + px + q$ 的图像。 写出该图像的方程，形式为 $y = x^2 + px + q$。 答案 ... 求出最低点的坐标。 答案 ( ..., ... )

1. Use the x-intercepts to write the quadratic in factorised form

The graph crosses the x-axis at $x=-4$ and $x=1.5$.

Since the coefficient of $x^2$ is $1$,
$y=(x+4)(x-1.5).$

2. Expand and simplify into the form $y=x^2+px+q$

$y=(x+4)(x-1.5)$
$y=x^2-1.5x+4x-6$
$y=x^2+2.5x-6.$

1. Complete the square for the quadratic found in part (b)(i)

From part (b)(i),
$y=x^2+\dfrac{5}{2}x-6$
$y=\left(x+\dfrac{5}{4}\right)^2-\left(\dfrac{5}{4}\right)^2-6$
$y=\left(x+\dfrac{5}{4}\right)^2-\dfrac{25}{16}-\dfrac{96}{16}$
$y=\left(x+\dfrac{5}{4}\right)^2-\dfrac{121}{16}$

2. Read off the minimum point from the completed square form

Since $\left(x+\dfrac{5}{4}\right)^2 \geq 0$, the minimum value occurs when
$x+\dfrac{5}{4}=0$
$x=-\dfrac{5}{4}$
Then
$y=-\dfrac{121}{16}$
Minimum point is
$\left(-\dfrac{5}{4},-\dfrac{121}{16}\right).$