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")#
![Tony B]()
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)#
