Vertex height form

Recall our earlier "square stack" graph:

This helped us understand the behavior of y=x^2, where the height is the square of the horizontal distance from the vertex. That sounds like a jumble, but all it means is \text{height }=\left(x\text{ units from vertex}\right)^2.

But now, imagine we slide it's lowest point from \left(0,0\right) to the coordinates of a new vertex \left(h,k\right).

Take a look at the graph below, and move around the vertex by dragging it. What do you notice? How does moving the vertex affect our "square stack" relationship?

As you adjust the vertex, notice that the entire square construction moves with it. The whole quadratic has the same shape and form relative to the shifted vertex.

This leads us to a key question: how can we mathematically express the quadratic in terms of the vertex coordinates?

First, notice that the number of squares in each stack still depends on their horizontal distance from the vertex, which is now located at x= h.

The distance from the vertex is equal to \left|x-h\right|. Remember that absolute value (\left|\cdots\right|) ensures we measure the horizontal distance regardless of whether the x value in question is to the left or right of h.

Since we know that \text{height }=\text{ (x units from vertex)}^2 , we can substitute in directly:

\text{height }=\left(\left|x-h\right|\right)^2=\left(x-h\right)^2

Note that squaring already guarantees the height is positive, so we can remove the absolute value. This formula for height accounts for any horizontal changes in the vertex, while maintaining the shape and form we saw graphically.

Additionally, when we vertically shift the vertex from 0 to k, we're just raising (or lowering) every point by the same amount, i.e.

\text{total height (y) }=\left(x-h\right)^2+k

Finally, there's one more element to consider: the scale factor a. Take a look at the graph below and notice how adjusting value of a affects the "square stack":

Notice that the vertex still remains in the same place as we scale the parabola by a, but the squares are "stretched" depending on the value of a.

There are two things to notice:

The squares are being stretched into rectangles, but the number of shapes in each column is not changing, it's still \left(x-h\right)^2. The height of each column is thus a\left(x-h\right)^2.
The vertex does not move - it's height above the x-axis is still k.

All of this comes together to create our finished form:

Foundations of Quadratics