Quadratic Graphing problem

Question

The vertex of the quadratic $x(x-2a)$ occurs when $x=4$.The vertex of the quadratic $(x+a)(x-3a)$ occurs when $x=L$. What is $L$?

Answer 1

$x(x- 2a)= x^2- 2ax= x^2- 2ax+ a^2- a^2= (x- a)^2- a^2$. If x= a, that is $-a^2$.0 For any other x it is $-a^2$ plus a positive number so larger than $-a^2$. The vertex is $(a, -a^2)$. We are told that the vertex is at x= 4 so a= 4. Then $(x+a)(x-3a)= (x+4)(x-12)= x^2- 8x- 48= x^2- 8x+ 16- 16+ 48= (x- 4)^2+ 32$. The vertex is again at x= 4.

Answer 2

The first coordinate of the vertex is the mean of the zeroes. In the first case that's $(0+2a)/2=a$ and we are told that this value is $4$. In the second case we have $(-a+3a)/2=a$ as well.