将 $x^{2} - 6x + 18$ 表示为 $(x - p)^{2} + q$ 的形式。由此，在下方坐标轴中画出 $y = x^{2} - 6x + 18$ 的图像草图。清晰地标出拐点（顶点）以及与坐标轴的所有交点。

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完整解法

1. Complete the square and simplify the remaining constant.

$x^{2}-6x+18=(x-3)^{2}-9+18=(x-3)^{2}+9$

答案

$x^{2}-6x+18=(x-3)^{2}+9$

1. Use the completed-square form to identify the turning point and shape of the parabola.

From part (a),
$y=x^2-6x+18=(x-3)^2+9.$
Since the coefficient of $(x-3)^2$ is positive, the graph opens upwards.

Turning point: $(3,9)$ .

2. Find and state the axial intercepts, then describe the required sketch.

For the $y$ -intercept, put $x=0$ :
$y=0^2-6(0)+18=18.$
So the $y$ -intercept is $(0,18)$ .

For the $x$ -intercepts, put $y=0$ :
$0=(x-3)^2+9.$
This has no real solution since $(x-3)^2\geq 0$ , so $(x-3)^2+9>0$ for all real $x$ .

Therefore, the graph lies above the $x$ -axis and has no $x$ -intercepts.

答案

Sketch an upward-opening parabola with turning point $(3,9)$ , crossing the $y$ -axis at $(0,18)$ , and with no $x$ -intercepts.

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