What is the vertex form of #y=x^2+3x+2#?

The vertex or turning point occurs at the point when the derivative of the function (slope) is zero.

#therefore dy/dx=0 iff 2x+3=0#

#iff x=-3/2#.

But #y(-3/2)=(-3/2)^2+3(-3/2)+2#

#=-1/4#.

Thus the vertex or turning point occurs at #(-3/2;-1/4)#.

The graph of the function verifies this fact.

graph{x^2+3x+2 [-10.54, 9.46, -2.245, 7.755]}

Given: #color(white)(....) y=x^2+3x+2#.....................(1)

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Consider just the #x^2+3x#

We are going to convert this to a 'perfect square' that is not quite equal to it. We then apply a mathematical 'adjustment' such that becomes equal to it.

#color(brown)("Step 1")#

Change the #x^2" to just "x#
Change the #3" in "3x" to " 1/2xx3=3/2#

Put it together in the form of #(x+3/2)^2#

As yet #(x+3/2)^2 # does not equal #x^2+2x# so we need to find out how to adjust it.

The adjustment is #(x^2+2x) -(x+3/2)^2#

#(x^2+2x)-(x^2+3x+9/4)#

So the adjustment is #-9/4#

#color(brown)("Note that the "+9/4" is an introduced value that is not wanted".)# #color(brown)("So we have to remove it; hence "-9/4)#

#(x^2+3x)=(x+3/2)^2-9/4#......................(2)

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#color(brown)("Step 2")#

Substitute (2) into equation (1) giving:

#y=(x+3/2)^2-9/4 +2#

#color(green)("Vertex Form "color(white)(...)->)color(white)(...)color(blue)(y=(x+3/2)^2 -1/4)#