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

Question 121913: Name the vertex for the parabol with equation
y= 3x^2-12x+16
Answer by jim_thompson5910(35256) (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:

From the equation we can see that a=3 and b=-12

Plug in b=-12 and a=3

Negate -12 to get 12

Multiply 2 and 3 to get 6

Reduce

So the axis of symmetry is

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

Start with the given polynomial

Plug in

Raise 2 to the second power to get 4

Multiply 3 by 4 to get 12

Multiply 12 by 2 to get 24

Now combine like terms

So the vertex is (2,4)

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