What is the vertex form of #y= (3-x)(3x-1)+11 #?

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Answer 1 · 2016-05-02T23:03:40

#y = -3(x-5/3)^2+49/3#

The vertex form of a quadratic equation is #y = a(x-h)^2+k#. In this form, we can see that the vertex is #(h, k)#.

To put the equation in vertex form, first we'll expand the equation, and then use a process called completing the square.

#y=(3-x)(3x-1)+11#

#=> y = -3x^2+9x+x-3+11#

#=> y = -3x^2+10x+8#

#=> y = -3(x^2-10/3x)+8#

#=> y = -3(x^2-10/3x+(5/3)^2-(5/3)^2)+8#

#=> y = -3(x^2-10/3x+25/9)+(-3)(-25/9)+8#

#=> y = -3(x-5/3)^2+49/3#

So, the vertex form is #y = -3(x-5/3)^2+49/3# and the vertex is #(5/3,49/3)#