Maths about graph

y=f(x)=ax^2+bx+c, vertex of the graph is at(3,0) and the y-intercept is 18.
This means that f(0)=18=>c=18
Rewrite f(x) as ax^2+bx+c=a(x^2+b/ax+c/a)=a[(x-b/2a)^2+c/a-b^2/4a^2]=a[(x-b/2a)^2+(b^2-4ac)/4a^2]
So (b^2-4ac)/4a=0 (since the vertex is (3,0) )
b^2=4ac=72a

a(x^2+b/ax+c/a)=a[(x-b/2a)^2+c/a-b^2/4a^2] E個 step 唔明, May you explain to me, thx so much~

2009-07-28 19:04:26 補充：
Could anyone explain me?Thx~ I don't why the ans. is that.