Sketch the graph of $y = 16 - (x + 3)^2$ on the axes below, indicating clearly the coordinates of the intercepts and the turning point on the graph.

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.