Question

using these three coordinates write an equation for the parabola in these forms

Answer 1

You know the parabola goes through those 3 points, so those points are solutions for the parabola, being a solution means that replacing the point will make the equation right (or true).
For your first expression
1. Point (0,0)
\[0=a(0-r_1)(0-r_2 ) / x=0 , y=0 \]
\[0=ar_1 r_2.............(I)\]
2. Point (20,12)
\[12=a(20-r_1)(20-r_2)=400a-20ar_2 -20ar_1+ar_1r_2\]
But you know from (I) that ar1r2=0
\[12=400a-20ar_2 -20ar_1.........(II)\]
3. Point (40,0)
\[0=a(40-r_1)(40-r_2)=1600a-40ar_2 -40ar_1+ar_1r_2\]
And again we know from (I) that ar1r2=0
\[0=1600a-40ar_2 -40ar_1\] dividing by 2
\[0=800a-20ar_2 -20ar_1........(III)\]

Now solving the equation system (II) and (III)
\[12=400a-20ar_2 -20ar_1.........(II)\]
\[0=800a-20ar_2 -20ar_1........(III)\]
we get a=-0.03, we also know for (I) that one of r1 & r2 should be 0, we pick one randomly and replacing a and r1=0 in (II) or (III) gives us the value of r2=40

So your first expression should be
\[y=-0.03(x)(x-40)\]

Now do the same for your second expression

Answer 2

parabola opens down a is negative vertex (20,12) line of symmetry x = 20
focus at (20, 6)