Question 35924
{{{y  =  x^ 2  - 4 x  +  3}}}


This is a parabola that opens up, since the coefficient of {{{x^2}}} is positive.  


You can complete the square or use a formula {{{x = -b/(2a) }}} to find the vertex and the line of symmetry.  


Completing the square looks like this:

{{{y  =  x^ 2  - 4 x  +  3}}}


Take half of the x coefficient (which is -2) and square it (which is 4), and add and subtract 4 from the right side of the equation.


{{{y  =  x^ 2  - 4 x  + _____+  3 - _____}}}
{{{y  =  x^ 2  - 4 x  + 4+  3 - 4}}}

{{{y = (x-2)^2 -1}}}


The line of symmetry will be the value of x that "zeros out the {{{(x-2)^2) }}}, which will be x=2.  This will also be the x coordinate of the vertex.  To find the y coordinate of the vertex, let x= 2, and you have y = -1.


Vertex = (2,-1)


The y intercept is where x = 0.  If x = 0, then y =3.


The x intercept is where y = 0.  If y = 0, then 
{{{0= x^2 - 4x+3}}}
{{{0 = (x-3)(x-1)}}}
So, the x intercepts will be at x =3 and x= 1.


Graph {{{graph(300,300, -4, 8, -4,10, x^2 -4x+3) }}}


R^2 at SCC