在下方坐标轴中画出 $y=(3-x)(x+2)$ 的函数图象草图。 清晰地标出图象与坐标轴交点以及曲线顶点的坐标。

1. Find the points where the graph crosses the $x$-axis.

At the $x$-axis, $y=0$.

$(3-x)(x+2)=0$

So

$3-x=0 \quad \text{or} \quad x+2=0$

$x=3 \quad \text{or} \quad x=-2$

The graph crosses the $x$-axis at $(3,0)$ and $(-2,0)$.

2. Find the point where the graph crosses the $y$-axis.

At the $y$-axis, $x=0$.

$y=(3-0)(0+2)=6$

The graph crosses the $y$-axis at $(0,6)$.

3. Find the turning point of the curve.

The roots are $x=-2$ and $x=3$, so the axis of symmetry is midway between them.

$x=\frac{-2+3}{2}=\frac12$

Substitute $x=\dfrac12$:

$y=\left(3-\frac12\right)\left(\frac12+2\right)$

$y=\frac52 \times \frac52=\frac{25}{4}$

Turning point is

$\left(\frac12,\frac{25}{4}\right).$

4. Determine the shape and sketch the graph using the key points.

Expand to identify the sign of the coefficient of $x^2$:

$y=(3-x)(x+2)=3x+6-x^2-2x=-x^2+x+6$

Since the coefficient of $x^2$ is negative, the parabola opens downwards.

Sketch a smooth downward-opening parabola through $(-2,0)$, $(3,0)$ and $(0,6)$, with maximum point $\left(\dfrac12,\dfrac{25}{4}\right)$.