在下方坐标轴上画出 $y=(x+4)(10-x)$ 的函数图象草图。
清晰地标出图象与坐标轴交点的坐标以及曲线上的最高点。

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

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

$(x+4)(10-x)=0$

So $x=-4$ or $x=10$.

The x-intercepts are $(-4,0)$ and $(10,0)$.

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

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

$y=(0+4)(10-0)=40$

The y-intercept is $(0,40)$.

3. Find the maximum point and sketch the curve.

The x-intercepts are $x=-4$ and $x=10$, so the axis of symmetry is halfway between them.

$x=\frac{-4+10}{2}=3$

When $x=3$,

$y=(3+4)(10-3)=7\times7=49$

Since the coefficient of $x^2$ is negative, the parabola opens downwards. Hence the maximum point is $(3,49)$.