Question

What is the value of a in this function’s equation?

Answer 1

a=-10/9

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