"Why = a(x  h)² + k?"  EVERYTHING ... ABOUT A QUADRATIC ...

Graphing by PointPlotting 
A graph is a portrait. It is a detailed picture of everything a rule does. For more information on what a graph is, see A Graph is A Portrait. Early in the study of algebra, Algebra I, the chief technique to produce a graph or portrait is pointplotting. For a short animating see PointPlotting Animation. In order to create a graph, a canvas or coordinate plane is required. Graph paper is available on the Planes and Masters Page.
A thorough examination of pointplotting work might be sufficient. Examine the work just below. 
Things You Should Notice  


Compare the info and the curve for each of the two above examples. See if you can determine information about the graph just by looking at the equation. It is important to be able to do this if you wish to avoid pointplotting. By now you should recognize a quadratic, with 2 as the highest degree in the equation, and know the curve it creates is a parabola. Examine the coefficient of the quadratic term since it determines the shape and concavity of the parabola. 

Use The 1, 3, 5, 7, 9, ... Shape to Reduce PointPlotting 
If the coefficient of the quadratic term is 1 or 1, the parabola is neither fat nor skinny but normal. The curvature on a normal parabola is a rise/fall of 1, then 3, then 5, then 7, then 9, then 11, then 13, and so on for each 1 unit it runs from the vertex. The steepness of the curve changes in exactly this way. If the coefficient of the quadratic term is 2 or 2, the curvature is a rise/fall of 2, then 6, then 10, then 14, then 18, for each unit it runs from the vertex  double the odd integers. The trick now is to find either the vertex or the xintercept so the graph may be drawn with little pointplotting. 
Graphing by Intercepts 
After learning the dilation from the coefficient of the quadratic term, find the xintercepts. Set the expression equal to 0 and solve for x to find the xintercepts of a graph. In many equations, this computation may be completed mentally.
On y=(x+1)² the xintercept is at 1. This is the vertex since only one value is on the horizontal line y=0. Draw the parabola with a normal dilation. On y=(x+3)(x+1), the xintercepts are at 3 and 1  2 units apart. The vertex is exactly half way between these numbers and 1 unit down at (2,1). On y=x(x+4), the xintercepts are at 0 and 4  4 units apart. The parabola is concave down with a normal dilation. Fit the parabola the the known points. 
Graphing by Inspection 
Sometimes it's easy to graph by inspection. Sometimes it's not. If the quadratic is written as y = a(x  h)² + k, it's easy. If the quadratic is written as y = ax² + bx + c, it's not as easy. Two "shifts" are stated in y = a(x  h)² + k. There is a " domain shifts" forced by the (x  h). It positions the curve left and right. There is a " range shifts" forced by the + k. It positions the curve up and down. See Composition of Functions and the how functions are created. See the graph of y = (x  2)²  4 from a composition of functions standpoint. See Curve Shifting and this graph and function explained in detail. In the curve y = (x  2)²  4, the squaring, symbolized by 1()² gives the curve its shape and dilation. The (x2) "recenters" the curve horizontally from a axis of symmetry of x=0 to an axis of symmetry of x2=0 or x=2. The  4 "shifts" the curve down 4. The graph y=(x3)²1 is shifted to the right 3 and down 1 compared to the curve y=x². The graph y=2(x2)²+1 is dilated and made steeper or "skinny" and shifted right 2 and up 1. The graph of y=(x3)²+4 is concave down with normal dilation, shifted to the right 3, and shifted up 4. 
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