What is the vertex form of #3y = 3x^2-4x+11#?

1 Answer
mason m
Dec 9, 2015

#y=(x-2/3)^2+29/9#

Explanation:

Vertex form of a quadratic equation: #y=a(x-h)^2+k#

The parabola's vertex is the point #(h,k)#.

First, divide everything by #3#.

#y=x^2-4/3x+11/3#

Complete the square using only the first #2# terms on the right. Balance the term you've added to complete the square by also subtracting it from the same side of the equation.

#y=(x^2-4/3xcolor(blue)+color(blue)(4/9))+11/3color(blue)-color(blue)(4/9#

#y=(x-2/3)^2+33/9-4/9#

#y=(x-2/3)^2+29/9#

From this, we can determine that the vertex of the parabola is at the point #(2/3,29/9)#.

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