f(x)=-3(x+2)²-3
y=-3[x²+4x+4)-3
y=-3x²-12x-12-3
y=-3x²-12x-15
f(x)= -3(x+2)² -3
y= -3(x²+4x+4)-3
y=-3x²-12x-15--->y=ax²+bx+c
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f(x)=-3(x+2)²-3
y=-3[x²+4x+4)-3
y=-3x²-12x-12-3
y=-3x²-12x-15
f(x)= -3(x+2)² -3
y= -3(x²+4x+4)-3
y=-3x²-12x-12-3
y=-3x²-12x-15--->y=ax²+bx+c