Sketch the graph of $y=(x+4)(10-x)$ 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. 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 $(3,49)$.