Sketch the graph of $y=-(x-2)^2+4$ on the axes below. Indicate clearly the coordinates of the points where the graph crosses the axes and the maximum point on the curve.

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:
$(0,0),\quad (4,0),\quad (2,4).$