Answer:
a=-10/9
Step-by-step explanation:
vertex=(0,10) (h,k)
y=a^2x+bx+c
y=c=10 when x=0
the x intercepts are (0,3) and (0,-3)
y=a (x-h)^2+k
y=a(x-0)+10
y=ax^2+10
find a at point (3,0)
y=a(3)^2+10
0=9a+10
9a=-10
a=-10/9
y=-10/9 x^2+10
check with points(-3,0)
0=-10(-3)^2/9+10
0=-90+90=0 correct
