we know that
the equation of a vertical parabola is equal to
y=a(x-x1)(x-x2)
where
a is the leading coefficient
x1 and x2 are the zeros of the function
In this problem we have
x1=-2
x2=6
substitute
y=a(x+2)(x-6)
If the parabola open upward, then the coefficient a must be positive
I will assume
a=1
so
y=(x+2)(x-6)
y=x^2-6x+2x-12
y=x^2-4x-12
using a graphing tool
