You have to combine these variables before you can do any thing else.
Given:" "y=(x-3)^2+7x-12 y=(x−3)2+7x−12
y=x^2-6x+9+7x-12y=x2−6x+9+7x−12
y=x^2+x-3y=x2+x−3...........................(1)
From this point you can 'complete the square' or do a part version of the process.
I am opting for the part process
Write as:
Consider the coefficient of +x+x in equation ( 1 ), which is 1
Apply (-1/2)xx1 = -1/2(−12)×1=−12
x_("vertex")=-1/2xvertex=−12....... Fast isn't it!
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color(brown)("A word of warning:")A word of warning: Given the standard form of
y=ax^2+bx+cy=ax2+bx+c you need to convert this to
y=a(x^2+b/ax)+cy=a(x2+bax)+c
In your case a=1a=1
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute x=-1/2x=−12 into equation (1)
y_("vertex")=(-1/2)^2+(-1/2)-3yvertex=(−12)2+(−12)−3
y_("vertex")=1/4-1/2-3 = -3 1/4yvertex=14−12−3=−314
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color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)Vertex→(x,y)→(−12,−314)