Graphing Quadratics

by Shelley Walsh ©2000

Here are the basic steps for learning how to graph quadratics.

  1. Learn what the graph of y=x2 looks like.
  2. Learn how to graph y=ax2 by stretching or shrinking y=x2 by a factor of a. A good guide for doing this is to plot (0,0), (1,a), and (-1,a)
  3. Graph y+ or -something=a(x+ or -something)2 by graphing y=ax2 translated over to where x+ or -something=0 and y+ or -something=0
  4. Be able to take any quadratic and write it in the form in step 3 by completing the square. (See  Quadratic Equations)
1. Make a chart in order to plot some points. Make sure to plot enough points to really get an idea of what it looks like. When graphing any equation it is a good idea to plot more points in places where something interesting is happening in the graph. In this case that would be around 0, because the direction is changing there. I've included 1/2 and 1/4 in my chart because this will help me know that the graph will be rounded rather than sharp at (0,0).


If you plot these points, you should get a graph that looks about like this.

The point (0,0), where it turns around, is called the vertex.

2. Stretching and Shrinking
To graph something of the form y=ax2 in a cleverer way than just plotting points it is helpful to think about it a bit and compare graphs of this sort to the basic y=x2 graph. To do this it is helpful to line up their charts and compare them.

The thing to notice here is that for the same x's the y's in the y=2x2 column are each twice as big as the ones the y=x2 column. The ones in the y=3x2 column are 3 times as big. In the y=(1/2)x2 column they are 1/2 the size. The effect of this geometrically will be that the graphs will be stretched or shrunk according to the size of the coefficient. In the next two columns they are negative, which will turn the graphs upside down. Here is what these graphs look like all graphed together.

A good guide to use when you graph equations of the form y=ax2 is to plot (0,0), (1,a), and (-1,a). This will both give you a start for showing the stretching effect and take care of the turning upside down part. Then sketch a curve through these points similar to the y=x2 curve, but stretched or shrunk by a factor of a.

3. Translating

By similar reasoning to step 2  with the chart comparisons, it is possible to see that if you add or subtract something to functions like these, it will raise or lower the graphs. For example, here is the graph of y=x2 -2.

What is a little harder to see is how translating to the right and left works. This is done by adding or subtracting something BEFORE the squaring, but it works backwards of the way you might think it would. If you add something it goes to the left and if you subtract something it goes to the right. So y=(x-2)2 looks like this
and y=(x+1)2 looks like this.
The reason for this is that in y=(x-2)2 you need larger x's to get the same y's as in y=x2 to compensate for the subtracting of the 2. In particular 2, the number that makes x-2 equal to 0, plays the same role as 0 in y=x2 , so what happens at (0,0) needs to get translated to (2,0). Similarly in y=(x+1)2, you don't need as large values for x to get the same y's and -1 corresonds to 0, so (0,0) gets translated to (-1,0). After learning about translating up and down, and right and left, you can put it all together and graph something like y=-2(x+2)2 +3. This should look like the graph of y=-2x2 translated 3 up and 2 to the left. Sometimes the directions of translation can be a bit confusing here with the vertical going the right way and the horizontal going the wrong way. I have found that for many people it helps, as silly as this might seem, to make them both go the wrong way, so at least they are consistent. You can do this by subtracting 3 from both sides of the equation to get y-3=-2(x+2)2. Then all you have to do to find the place that corresponds to (0,0), the vertex, is to set x+2 and y-3 equal to 0. When you do that, you get a graph that looks like this.

There is another way to interpret this that is also useful for lots of other graphing (See  Graphing the Sine and Cosine Functions ), which is to change your coordinates to a translated coordinate system. This may sound difficult and scary, but it really isn't. All you have to do is make a substitution, introducing two new variables X and Y  to represent the coordinates of your translated coordinate system. You do this in such a way that the equation becoming one like those in step 2. In the above example this means that you could let X=x+2 and Y=y-3. Then the equation becomes Y=-2X2. Then the only thing you have to do is to find out where this X,Y coordinate system is relative to the x,y coordinate system, but that's not too difficult either, because to know where a coordinate system is all you have to know is where (0,0) is. To do that, set X and Y equal to zero. X=x+2 and Y=y-3, so that means that the origin for the X,Y coordinate system is where x+2=0 and y-3=0. Solve these equations and you get (-2,3) for the origin of the X,Y coordinate system. Then to graph y-3=-2(x+2)2 or Y=-2X2 in the X,Y system, draw a dotted coordinate axis through (-2,3) and then graph Y=-2X2 using that as your coordinate axis and totally ignoring the x,y coordinate axes.

To get an good guide for the shape of your graph, just like in step 2, you can plot (0,0), (1,a), and (-1,a) (in this case (0,0), (1,-2), and (-1,-2), since a=-2), but relative to the X,Y axes. This in effect means that once you find your vertex, you plot the vertex and a point 1 to the right and a up (down if a is negative) and a point 1 to the left and a up (down if a is negative). So in this example that is 1 to the right 2 down, 1 to the left 2 down from the vertex (-2,3).

4. Completing the Square

Believe or not, step 3  really allows you to graph all possible quadratics, because you can use the method of complete the squares  to get any quadratic into that form. Here's how to do it. Suppose you want to graph

y=2x2-4x+5.

This is a little bit trickier than completing the square for solving the equations, because you don't have it set equal to 0, so you can't get rid of the 2 so easily. There are a number of ways of dealing with this, but this is the one I think is the most dependable for keeping track of things. Instead of dividing both sides by the 2, you factor it out, but it turns out to be most convient to just factor it out of the first two terms. When you do that you will get

y=2(x2-2x                 )+5.

I've made this one easy by making the coefficient on the second term a multiple of two also, but if it wasn't you could just make it factor out by dividing by it and putting up with the fraction. I have also left a space to leave room for the completing of the square. To complete the square, divide 2 by 2 and square it to get 1, and this time instead of adding it to both sides, add it and subtract it inside the parentheses like this.

y=2(x2-2x+1-1)+5

I know this sounds daft, because it doesn't seem like it does anything to add and then subtract, but by doing this you can divide the expression in two pieces in the right way for what we want, and by adding and subtracting inside the parentheses it keeps track of the fact that we are really not adding 1, but 2, because the 1 is inside the parentheses, so it is getting multiplied by 2. Then you think of the inside of the parentheses as divided in two pieces like this

y=2([x2-2x+1]+[-1])+5

The first piece is the perfect square and the second is the remaining bit. Distribute the 2 into each piece, merely writing the multiplication for the first piece, and doing it for the second piece like this.

y=2(x2-2x+1)+-2+5

Now write the perfect square as the perfect square that it is and add the -2 and the 5 to get

y=2(x-1)2+3

Now you can apply the techniques from step 3  to graph it and you should get a graph that looks like this.



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