Answer:
(-2 -2sqrt(2/3),0)
(-2 +2sqrt(2/3),0)
Explanation:
The vertex form of a parabola is
y= a(x-h)^2 +k
where (h,k) is the vertex
y = a(x--2)^2 -8
y = a(x+2)^2 -8
We can substitute the other point in to find a
4 = a(0+2)^2 -8
4 = a(2)^2 -8
4 = 4a-8
Add 8 to each side
4+8 = 4a-8+8
12 = 4a
Divide by 4
12/4 =4a/4
a=3
y = 3(x+2)^2 -8
To find the x intercepts, set y =0 and solve for x
0 = 3(x+2)^2 -8
Add 8 to each side
8 =3(x+2)^2 -8+8
8 =3(x+2)^2
Divide by 3
8/3 =3/3(x+2)^2
Take the square root of each side
±sqrt(8/3) = sqrt((x+2)^2)
±sqrt(8/3) = (x+2)
We know that sqrt(ab) = sqrt(a)sqrt(b)
±sqrt(4) sqrt(2/3) = (x+2)
±2 sqrt(2/3) = (x+2)
Subtract 2 from each side
-2 ±2sqrt(2/3) = (x+2)-2
-2 ±2sqrt(2/3) = x
(-2 -2sqrt(2/3),0)
(-2 +2sqrt(2/3),0)