There are three things we need to consider as a pre-amble before we start.

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#color(blue)("Point 1")#

Consider #(3x)^2# Inside the brackets the coefficient is presented as 3. Outside the bracket it has been squared so it will be 9 in that:

#9xx(x)^2=(3x)^2 # another example #->" "16xx(x)^2=(4x)^2#

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#color(blue)("Point 2")#

#1/3xx(3x-15)^2 =((3x)/(sqrt(3))-15/sqrt(3))^2#

so #1/9(3x-15)^2=((3x)/3-15/3)^2#

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#color(blue)("Point 3")#

To convert the given equation into vertex form we need to end up with the format of:

#y=a(x-b/(2a))^2 +c" "# where #b# can be positive or negative.

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#color(blue)("Solving your question")#

With the format of the given question you are already part way to building the vertex equation format of completing the square. So this is what I am going to do.

Given:#" " y= (1/6)(3x-15)^2-31#

To remove the coefficient of #x# within the brackets multiply the bracketed part by 1, but in the form of #color(blue)(9/9)#

#y= color(blue)(9/9)(1/6)(3x-15)^2-31#

#y=(color(blue)(9))/6((3x)/(color(blue)(3))-15/(color(blue)(3)))^2-31#

#y=9/6(x-5)^2-31" "color(brown)("This is vertex form")#

Thus:

#x_("vertex")=(-1)xx(-5)=5#

#y_("vertex") = -31 # Notice that this is the value of the constant #c#

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Vertex#" "=" "(x,y)" "->" "(5,-31)#