在下方坐标系中画出 $y=-(x-2)^2+4$ 的函数图像草图。 清晰地标出图像与坐标轴交点的坐标以及曲线上的最高点。

1. Identify the maximum point and shape of the parabola

The graph is given in completed-square form:
$y=-(x-2)^2+4.$
Since the coefficient of $(x-2)^2$ is negative, the parabola opens downwards.

The maximum point is
$(2,4).$

2. Find the intercepts with the axes

For the $y$-intercept, put $x=0$:
$y=-(0-2)^2+4=-4+4=0.$
So the graph crosses the $y$-axis at $(0,0)$.

For the $x$-intercepts, put $y=0$:
$0=-(x-2)^2+4$
$(x-2)^2=4$
$x-2=\pm 2$
$x=0 \text{ or } x=4.$
So the graph crosses the $x$-axis at $(0,0)$ and $(4,0)$.

3. Sketch and label the required graph clearly

Draw a smooth downward-opening parabola through $(0,0)$ and $(4,0)$, with its highest point at $(2,4)$.

Label the intercepts and maximum point clearly:
$(0,0),\quad (4,0),\quad (2,4).$