Question 35606
You will need to complete the square to write this in the correct form.

{{{y = x^2 - 6x + 8}}}


Take half of the x coefficient and square it.  Next, add AND subtract it from the same side.  In this way you will NOT change the equation:

{{{y = x^2 - 6x + ____ - _____ + 8}}}


Half of -6 is -3, and (-3) squared will be 9, so add and subtract 9 from the right side of the equation:
{{{y = x^2 - 6x + ____ - _____ + 8}}}
{{{y = x^2 - 6x + 9 - 9 + 8}}}


This forms a perfect square trinomial in the first three terms on the right side:
{{{y = (x-3)^2 -1}}}


So a = 1, h= 3, and k= -1.


The line of symmetry is x= 3.


The graph is a parabola with vertex at (3,-1).  

{{{graph(300,300, -10,10,-10,10, x^2-6x+8) }}}


Compared to the graph of y = x^2, this graph is exactly the same shape.  The vertex is just moved from (0,0) to (3,-1).  Here is a comparision:
{{{graph(300,300, -10,10,-10,10, x^2-6x+8, x^2) }}}


R^2 at SCC