Question 108823
{{{y=x^2-5x+3}}}

Since {{{b = -5}}}, {{{-b}}} will be {{{5}}}

Then

Axis of symmetry: {{{x = -b/2a}}}    …..=>…… 

Axis of Symmetry: {{{x = 5/2 = 2(1/2)}}}   ……or use decimal…{{{x = 2.5}}}

This is {{{one}}} point through what the axis of symmetry passes. You need one more point to draw that line, and that point has to be a {{{vertex}}}. Substitute {{{5/2}}} for {{{x}}} in the {{{y = x^2 -5x + 3}}} and find {{{y}}}.

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

{{{y = 6.25 - 12.5 + 3}}}

{{{y =-3.25}}}

So, the vertex is: ({{{2.5}}},{{{-3.25}}})

The graph will be a curve called a parabola

Here is the table:
 x  | y
---------
  0 | 3
  1 |-1
  2.5 |-3.25
  3 |-3
  4 |-1
  5 | 3
  6 | 9






*[invoke plot_any_graph "x^2-5x+3", -7, 7, -5, 5, 600, 600]