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
