Draw a picture first, so that DF is perpendicular to BC, and point F passes through point D..

It is known that CF=3, trapezoidal height DF = 4,0.

Let's discuss the simplest one first: 5

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Discussion 19

The height of each stage has been set, which is easy to follow. The base AE=4 of the triangle is 4.

0<= X< is at 5 o'clock, and Y=(4*4X/5)/2=8X/5.

5<= X< When = 19, Y=AE*h/2=4*4/2=8.

19 & lt； X & lt=24，Y=[4*4*(24-X)/5]/2=8(24-X)/5。

The function has been written, and the image is easy to draw. Draw it yourself. Divide it into three sections.