1) The number beside `x^2` is A, so:

`f(x)=3(x+B)^2+C`

2) Factor out A. That is, divide everything by 3:

`f(x)=3(x^2+2x+7/3)`

3) The number that is next to `x` is double B

So, to find B, take 2 and cut in half

So B =1

`f(x)=3(x+1)^2+C`

4) There are a few ways to find C. Here is one of them:

You want

``

``

``

``

``

`4=C`

So:

`f(x)=3(x+1)^2+4`

One of the reasons this is useful, is it immediately tells you the minimum of the parabola. (If A is negative, it is an upside-down parabola, and instead the parabola has a maximum)

The minimum value occurs at `x=-B`

So, for this problem, the minimum occurs at `x=-1`

Plug in to find what the minimum value is:

`f(-1)=3(-1+1)^2+4=0+4=4`

That is, the minimum value is just C

To summarize:

If `f(x)=A(x+B)^2+C`

then the minimum occurs at `x=-B` , and the minimum is `C`

(if A is negative, then the maximum occurs at -B, and the maximum is C)

So, for us, the value of `x` for which `f(x)` is a minimum is -1

and the minimum value of `f(x)` is 4

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