Question 133422
You can tell whether a parabola opens 'up' or 'down' by the sign on the high order term. In this case that term is {{{x^2}}}. Since the sign is positive, the parabola opens 'up'.

To determine if the entire parabola lies above the x axis, you can set y = 0 and then solve for x. If there is no value of x that allows a y =0, then the entire parabola is above the x axis. If there is a single value of x that makes y=0, then the parabola "just touches" the x axis. If there are 2 values that make y=0, then the parabola crosses the x axis and has at least some part of below the x axis. 

That is the case here:
{{{y = x^2 -4x + 3}}}
{{{y = (x-3)(x-1)}}}
Setting y =0 and solving for x yields values of x at 1 and 3 that allow y to be 0. --> two x values imply some part is below 


{{{graph (600,400, -10, 10, -10, 10, x^2-4x+3 ) }}}