Determine the equation of a cubic function with zeros 2 √3 and -1 whose graph passes through the point (2, -18)

Answer 1

y = a(x-2)(x-√3)(x+1)
since y(2) = -18, this is impossible. So maybe you meant the zeroes are
2±√3 and -1. In that case,
y = a(x-(2-√3))(x-(2+√3))(x+1) = a(x^3 - 3x^2 - 3x + 1)
since y(2) = -18, that means that
a(-9) = -18
so a = 2 and y = 2(x^3 - 3x^2 - 3x + 1)