SOLUTION: Name the vertex for the parabol with equation y= 3x^2-12x+16

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Question 121913: Name the vertex for the parabol with equation
y= 3x^2-12x+16
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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%2F%282a%29

From the equation y=3x%5E2-12x%2B16 we can see that a=3 and b=-12

x=%28--12%29%2F%282%2A3%29 Plug in b=-12 and a=3


x=12%2F%282%2A3%29 Negate -12 to get 12


x=%2812%29%2F6 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%5E2-12x%2B16 Start with the given polynomial


y=3%282%29%5E2-12%282%29%2B16 Plug in x=2


y=3%284%29-12%282%29%2B16 Raise 2 to the second power to get 4


y=12-12%282%29%2B16 Multiply 3 by 4 to get 12


y=12-24%2B16 Multiply 12 by 2 to get 24


y=4 Now combine like terms


So the vertex is (2,4)