SOLUTION: Graph y= -1/3(x+6)^2+3

Hi

the vertex form of a parabola,  where(h,k) is the vertex

y= -1/3(x+6)^2+3  

vertex is Pt(-6,3) negative 'a' of -(1/3) tells us the parabola opens downward

y intercept (when x = 0)

y = -1/3(6)^2 + 3

y = -12 + 3 = -9  Pt(0,-9)

x intercepts (when y = 0)

0 = -1/3(x+6)^2+3  

0 = -(x+6)^2 + 9

(x+6)^2 = 9

(x+6) = ±3

x +6 = 3    x = -3  Pt(-3,0)

x +6 = -3   x = -9  Pt(-9,0)