To begin, we know the standard form of a parabola is y = a(x - h)^2 + k, where (h, k) is the vertex. Since we have two x-intercepts, we can set up the following equations using the roots:
(x + 10)(x + 2) = 0
x^2 + 12x + 20 = 0
Solving for x using the quadratic formula, we get:
x = (-12 Β± β(12^2 - 4(1)(20))) / (2(1))
x = (-12 Β± β(64)) / 2
x = -6 Β± 4
So the x-intercepts are (-10,0) and (-2,0). To find the vertex, we can use the axis of symmetry, which is the line midway between the x-intercepts:
axis of symmetry = (10 + 2) / 2
axis of symmetry = 6
Therefore, the vertex is (6,-4). Plugging this into our standard form equation, we get:
y = a(x - 6)^2 - 4
To find the value of a, we can use either of the x-intercepts:
0 = a(-10 - 6)^2 - 4
40a = 4
a = 1/10
Therefore, the equation of the parabola is:
y = (1/10)(x - 6)^2 - 4