Question

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

Answer 1

`(-3/2;-1/4)`

Explanation:

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]}

Answer 2

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

Explanation:

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`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)`