Question 89934
<pre><font size = 5><b>
Identify the axis of symmetry, create a 
suitable table of values,and sketch the 
graph including the axis of symmetry. 

y = {{{x^2-5x+3}}}

Make this table:
 x  | y
---------
 -1 | 9 
  0 | 3
  1 |-1
  2 |-3
  3 |-3
  4 |-1
  5 | 3
  6 | 9

Plot those 9 points:

{{{drawing(400,400,-2, 7,-5,4,

graph(400,400,-2, 7,-5,4),

locate(-1-.08,9+.22,o), locate(0-.08,3+.22,o), locate(1-.08,-1+.22,o), locate(2-.08,-3+.22,o),
 locate(3-.08,-3+.22,o), locate(4-.08,-1+.22,o), locate(5-.08,3+.22,o), locate(6-.08,9+.22,o)) }}}

Now draw a smooth U-shaped curve through them. 
Even though those bottom two points are side by
side, make sure you draw the curve down below
the first one and then back up through the
second one.

This curve is called a parabola. 

{{{drawing(400,400,-2, 7,-5,4,

graph(400,400,-2, 7,-5,4,x^2-5x+3),

locate(-1-.08,9+.22,o), locate(0-.08,3+.22,o), locate(1-.08,-1+.22,o), locate(2-.08,-3+.22,o),
 locate(3-.08,-3+.22,o), locate(4-.08,-1+.22,o), locate(5-.08,3+.22,o), locate(6-.08,9+.22,o)) }}}

Now let's get the axis of symmetry.

Rule:
The axis of symmetry of the graph of

y = {{{ax^2+bx+c}}}

is the vertical line whose equation is x = {{{-b/(2a)}}}

In your problem

y = {{{x^2-5x+3}}}

a = 1, b = -5 and c = 3

So the axis of symmetry is the vertical 
line whose equation is

x = {{{-b/(2a)}}}

x = {{{-(-5)/2(1)}}}

x = {{{5/2}}}

or as a decimal:

x = 2.5

It's the green vertical line below:

{{{drawing(400,400,-2, 7,-5,4,

graph(400,400,-2, 7,-5,4,x^2-5x+3,999*(x-2.5)),

locate(-1-.08,9+.22,o), locate(0-.08,3+.22,o), locate(1-.08,-1+.22,o), locate(2-.08,-3+.22,o),
 locate(3-.08,-3+.22,o), locate(4-.08,-1+.22,o), locate(5-.08,3+.22,o), locate(6-.08,9+.22,o)) }}} 

You weren't asked for the vertex, the point at 
the very bottom of this parabola, but you can
find it by solving the system of equations consisting
of the equation of the parabola and the equation of
the axis of symmetry:

y = {{{x^2-5x+3}}}
x = 2.5

Substituting 2.5 for x in the first equation

y = {{{(2.5)^2-5(2.5)+3}}}
y = {{{-3.25}}}

So the vertex is the point (2.5,-3.25)

{{{drawing(400,400,-2, 7,-5,4,

graph(400,400,-2, 7,-5,4,x^2-5x+3,999*(x-2.5)),
locate(2,-3.5,"VERTEX(2.5,-3.25)"),
locate(-1-.08,9+.22,o), locate(0-.08,3+.22,o), locate(1-.08,-1+.22,o), locate(2-.08,-3+.22,o),
locate(2.5-.08,-3.25+.22,o), locate(3-.08,-3+.22,o), locate(4-.08,-1+.22,o), locate(5-.08,3+.22,o), locate(6-.08,9+.22,o)) }}}
Edwin</pre>