How do i write y=-2x^2 16x-7 in vertex form

To do this you need to "complete the square" as follows: 

y = 2x^2 - 12x + 16 
y = 2(x^2 - 6x + 8) 
y = 2(x^2 - 6x +8 +1 -1) (you "complete the square" by bringing the last term (8) up to "half of the middle co-efficient squared". Middle co-efficient is -6, half is -3, squared is 9. To get from 8 to 9 you must add 1 (+1), but if you add 1 you must subtract 1 (-1) so that you don't change the overall value of inside the bracket) 

y = 2(x^2 - 6x +9) - 2 (bring the -1 outside of the brackets) 
y = 2(x-3)^2 - 2 

and that is your final answer. I hope this helps!

B. 16is the absolute value