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)
