Question 121913
To find the vertex, we first need to find the axis of symmetry (ie the x-coordinate of the vertex)

To find the axis of symmetry, use this formula:


{{{x=-b/(2a)}}}


From the equation {{{y=3x^2-12x+16}}} we can see that a=3 and b=-12


{{{x=(--12)/(2*3)}}} Plug in b=-12 and a=3



{{{x=12/(2*3)}}} Negate -12 to get 12



{{{x=(12)/6}}} Multiply 2 and 3 to get 6




{{{x=2}}} Reduce



So the axis of symmetry is  {{{x=2}}}



So the x-coordinate of the vertex is {{{x=2}}}. Lets plug this into the equation to find the y-coordinate of the vertex.




{{{y=3x^2-12x+16}}} Start with the given polynomial



{{{y=3(2)^2-12(2)+16}}} Plug in {{{x=2}}}



{{{y=3(4)-12(2)+16}}} Raise 2 to the second power to get 4



{{{y=12-12(2)+16}}} Multiply 3 by 4 to get 12



{{{y=12-24+16}}} Multiply 12 by 2 to get 24



{{{y=4}}} Now combine like terms



So the vertex is (2,4)