在下方坐标系中画出 $y = 16 - (x + 3)^2$ 的函数图象草图，并清晰地标出图象上截距和顶点的坐标。

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完整解法

1. Identify the turning point and shape of the parabola

$y=16-(x+3)^2=-(x+3)^2+16$
This is of the form $y=-(x-p)^2+q$ , so the turning point is $(-3,16)$ .

Since the coefficient of the squared term is negative, the parabola opens downwards.

2. Find the x-intercepts

At the x-intercepts, $y=0$ .
$16-(x+3)^2=0$
$(x+3)^2=16$
$x+3=\pm 4$
$x=1 \text{ or } x=-7$
So the x-intercepts are $(1,0)$ and $(-7,0)$ .

3. Find the y-intercept and complete the sketch

At the y-intercept, $x=0$ .
$y=16-(0+3)^2$
$y=16-9=7$
So the y-intercept is $(0,7)$ .

Sketch the curve passing through $(-7,0)$ , $(1,0)$ and $(0,7)$ , with maximum turning point $(-3,16)$ , opening downwards.

答案

Sketch a downward-opening parabola with turning point $(-3,16)$ , x-intercepts $(-7,0)$ and $(1,0)$ , and y-intercept $(0,7)$ .

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