Identify the axis of symmetry, create a suitable 
table of values, then sketch the graph 
(including the axis of symmetry).

y = –x² + 3x – 3

For the quadratic equation 

y = Ax² + Bx + C the axis of symmetry has the 
equation

x = -B/(2A)

In your problem,

y = –x² + 3x – 3,

A = -1, B = 3, C = -3, so the axis of symmetry has
the equation

x = -B/[2A]

x = -(3)/[2(-1)]

x = 3/2

First let's graph the axis of symmetry.  We have 
just one choice for x, namely 3/2.  We just choose 
arbitrary values for y, always choosing 3/2 for x, 
and draw the graph of the the axis of symmetry:

AXIS OF SYMMETRY

 x  |  y |   (x,y)  |
--------------------|
3/2 |  5 | (3/2, 5) |     
3/2 |  3 | (3/2, 3) |       
3/2 | -2 | (3/2,-2) |       
3/2 |  6 | (3/2, 6) |        
3/2 | -1 | (3/2, 1) |         
3/2 |  4 | (3/2, 4) |

Those values for y are completely arbitrary!
Plotting these points and drawing through them
we get this graph, a vertical line
(Notice that 3/2 is just 1 and a half).



Now we draw a graph of your quadratic equations:

y = –x² + 3x – 3

QUADRATIC EQUATION

 x | y  |   (x,y)  |
--------------------|
-1 | -7 | (-1,-7) |     
 0 | -3 |  (0,-3) |       
 1 | -1 |  (1,-1) |       
 2 | -1 |  (2,-1) |        
 3 | -3 |  (3,-3) |         
 4 | -7 |  (4,-7) |        

Now we plot those points and draw a
smooth (green) curve through them:



You will notice that the graph of the axis of 
symmetry, which is the red vertical line, 
bisects the green graph into two symmetrical 
halves.  

Edwin McCravy