There are 2 options to solve that. 1. The first one is by derivatives. f(x)=x^2+12x+36 f'(x)=2x+12 then you solve that for f'(x)=0 0=2x+12 x=(-6) you have x so for (-6) solve the first equation, then you find y y=(-6)^2+12*(-6)+36=(-72) so the vertex is (-6, -72) 2. The second option is to solve that by equations: for x we have: x=(-b)/2a for that task we have b=12 a=1 x=(-12)/2=(-6) you have x so put x into the main equation y=(-6)^2+12*(-6)+36=(-72) and we have the same solution: vertex is (-6, -72)
For next task, I will use the second option: y=x^2-6x x=(-b)/2a for that task we have b=(-6) a=1 x=(6)/2=3 you have x so put x into the main equation y=3^2+(-6)*3=(--9) and we have the same solution: vertex is (3, -9)
Answer and workings in the attachment below. Also, do me one big favour. Let me know if it's the answer your teacher or text book ends up giving you. That would give me some satisfaction.