Vertex is 4, 0
vertex form = y = a (x-4)^2 + 0 sub in the given point (6,2) to calculate 'a'
2 = a ( 6-4)^2
a = 1/2
your equation is then y = 1/2 (x-4)^2 + 0
General Equation of Parabola: $y=a(x-h)^2 + k$ where $(h,k)$ is the vertex of the parabola.
We have $y=a(x-4)^2 + 0$
We can’t the vertex $(4,0)$ again to plug in for $(x,y)$ because it will yield: $0=0+0.$
Let’s use the other point given(namely $(6,2)$): $2=a(6-4)^2 + 0$
We have $a=\frac{2}{4}=\frac{1}{2}.$
Thus the equation is $y=\frac{1}{2}(x-4)^2$