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A function problem in eighth grade mathematics
(1) Point P is perpendicular to the Y axis, and S(CPO)=(4*4)/2=8.

(2)A(x, 0) is (-x*4)/2+8= 16, so A(-4, 0).

S(PAO)=(m*A0)/2= 16,∴m=8

(3) S△BOP=2S△AOP, △AOP= 16, so S△BOP=32=(m*BO)/2(S= bottom multiplied by height divided by 2).

∴BO=8, pay attention to point P (4, m), m=8,

∴p(4,8) and b (8 8,0) determine a straight line at the same time. At this time, the straight line is determined, and both of them are on BD.

You can set this straight line as y=kx+b, then these two points can be substituted into this straight line.

8k+b=0 and 4k+b=8, so the solution is k=-2 and b= 16, so BD:y=-2x+ 16.