hard!

2 (x-4)^2 +8      expand to find  a    b   c

2x^2 -16x + 40           a = 2    b = -16   c = 40

then    3a = 6    3b = -48    3c = 120

6x^2 -48x +  120

6 (x-4)^2  +24                 h = -4

Which makes sense, because all you did was multiply the original expression by 3

3 * (2(x-4)^2 + 8) =    6(x-4)^2+24     h = -4

That makes sense. Thanks a lot!

\(ax^2+bx+c=2(x-4)^2+8\\~\\ 3(ax^2+bx+c)=3[2(x-4)^2+8]\\ 3(ax^2+bx+c)=6(x- {\color{red} 4})^2+24\\ \)

SO

h=4

LaTex:

ax^2+bx+c=2(x-4)^2+8\\~\\
3(ax^2+bx+c)=3[2(x-4)^2+8]\\
3(ax^2+bx+c)=6(x -{\color{red} 4})^2+24\\

Yes! That is what I put at first, too, before I made a calculation error and fixed it to 3