This is a quadratic question of y=a(x-h)^2+k. And i don t understand how to find a.?

when you say that the graph starts at (0, 0) and ends at (10, 0) do you mean both points are on the graph. If yes, you could use either of the points to find a value. Since the graph is symmetrical around x=5, then the value at x=10 and x=0 should be the same.

Plug a point back into your equation.

y= a(x-5)^2+4
plug in 10, 0 x=10 y=0
0 = a(10-5)^2+4
0= 25a+4
25a=-4
a=-4/25 = -0.16
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back check
y= -4/25*(x-5)^2+4
0= -4/25*(10-5)^2+4
0=-4+4
but does (0,0 )
0 = -4/25(0-5)^2+4