Question #35cd4

You have not fully explained the situation from which you wish to obtain #-b/(2a)#. However, I think I know what you are after.

It is part of the process called 'completing the square' so I will 'walk you through' the process.

Consider the standardised equation format of #y=ax^2+bx+c#
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#color(blue)("The process of completing the square")#

Factor out a from part of it:

write as #y=a(x^2+b/a x)+c#

#color(brown)("................................................................")#
#color(brown)("explaining why we divide by "a" in "b/ax")#
To return this into the original equation form you would multiply EVERYTHING inside the bracket by #a#

#axxx^2=ax^2# which matches the original

#axxb/ax" "# which is the same as #" "a/axxbx#
This in turn is the same as #1xxbx=bx# which matches the original equation.
#color(brown)("................................................................")#

Take the power of 2 from #x^2# and put it outside the brackets
However, the changes in what follows introduces an error. So we compensate for this error by introducing the correction facto of #k#

#y=a(x+b/ax)^2+k+c#

Remove the #x# from #b/ax#

#y=a(x+b/a)^2+k+c#

All you need to do now if determine the value of #k#
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#color(blue)("Now comes your question part")#

The vertex (tip of the rounded bottom or top of the curve) as an #x# component and a #y# component

#x_("vertex")=color(red)(-1/2)xxb/a# this is the same as #-b/(2a)#

You just have to know that the #color(red)(-1/2xx)# is what is always needed to make the system work. You always have to do it

Substitute values and you have almost got the whole thing.
There is a bit more manipulation to determine the value of #k# but you have not asked about that.
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#color(blue)("An example")#

Suppose that we were given the values #y=3x^2+2x-5#

#a=+3"; "b=+2#

so #-b/(2a)=-((+2)/(2xx3)) = -1/3#