how to find the quadratic equation in standard form for the parabola with the roots and vertex

Question

Determine the quadratic equation in standard form for the parabola with zeros -7 and 1 and vertex (-3, 32).

I have this so far but i'm not sure where to go...

(x-7)(x+1)=0

x^2 - 6x - 7

y=a(x^2-6x-7)

32=a(-3^2-6(-3)-7)

32=a(-9+18-7)

a=16

Answer 1

a(x+7)(x-1) = a(x^2 +6x -7)

vertex (-3,32)

32 = a(9-18-7) = -16a

a = -2

y = -2(x^2 +6x -7)

y = -2x^2 -12x +14 in standard form

y = -2(x+3)^2 +32 in vertex form

y = -2(x+7)(x-6) in factored form