Sketch the graph of $y=(3-x)(x+2)$ on the axes below. Indicate clearly the coordinates of the points where the graph crosses the axes and the turning point of the curve.

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 $\left(\dfrac12,\dfrac{25}{4}\right)$.