b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.

Before you talk about completing the square, remind your students about the perfect square trinomial. In general form, a perfect square trinomial is a quadratic of the form m² ± 2mn + n², which can be factored into the form (m ± n)(m ± n) = (m ± n)².

Notice that half of the middle term is equal to the square root of the first term times the square root of the last term. This is a requirement for a quadratic to be a perfect square trinomial. Not all quadratics are in this form. Some algebraic manipulation can be used to put any quadratic equation into a form where one side is a perfect square trinomial. That's called completing the square.

What about the equation y = x² – 4x + 3? This quadratic is factorable as (x – 1)(x – 3). Clearly not a perfect square trinomial, but no one's perfect, right?

First, subtract 3 from both sides to get y – 3 = x² – 4x. Now, we can add whatever we want to the right hand side, as long as we do exactly the same thing to the left hand side. We want a 4 where the 3 used to be (because 4 is a perfect square of 2), so we get y – 3 + 4 = x² – 4x + 4. The right hand side is now in the proper perfect square trinomial form, and can be factored as (x – 2)². Our equation is then y + 1 = (x – 2)².

For every x-value we plug into the right hand side of the equation, there is a corresponding y-value on the left hand side. When x = -2, y must be equal to 15, for example. Notice that the right hand side of the equation is never negative, since we are squaring a number, and that always produces a positive number.

The smallest thing the right hand side can possibly be is 0 (when x = 2). This is going to translate to the smallest possible value that y can be (y + 1 = 0, which means y = -1). In mathematical terms, that point represents the minimum of the function. In this case, the point (2, -1) represents the minimum point of our parabola, which is called the vertex.

Every quadratic equation has either a minimum or a maximum, depending on whether it points up or down (whether it looks like a happy face or a sad face). Notice that we could not have easily found the coordinates of the vertex from y = (x – 1)(x – 3). We had to put the quadratic in a different form that highlighted the coordinates of the vertex in order to extract this information.