[tex]y=ax^2+bx+c[/tex]
[tex]f(x)=-x^2+3[/tex]
a = -1 < 0 therefore the parabola open down.
The function is increasing on the interval [-∞, v] and decreasing on the interval [v, ∞].
The parabola of [tex]y=-x^2[/tex] is shifted 3 units up, therefore the vertex is in (0, 3).
[tex]f\nearrow\ for\ x\in[-10,\ 0]\\\\f\searrow\ for\ x\in[0,\ 10][/tex]
Answer:
A = -10
B = 0
C = 0
D = 10
