Question 19436
You can find the x-coordinate of the vertex of this parabola from: {{{x = -b/2a}}}
{{{x = -(-4)/2}}} = 2
The y-coordinate is found by substituting this value of x (2) into the quadratic equation and solving for y:
{{{y = 2^2 - 4(2) + 1}}}
{{{y = 4 - 8 + 1}}}
{{{y = -3}}} 

The vertex is at (2, -3)

Then you could find the x-intercepts (there will be two of 'em) by letting y = 0 and solving for the roots.

{{{x^2 - 4x + 1 = 0}}} Use the quadratic formula to solve: {{{x = (-b+-sqrt(b^2-4ac))/2a}}}

{{{x = (-(-4)+-sqrt((-4)^2 - 4(1)(1)))/2(1)}}}
{{{x = (4+-sqrt(12))/2}}}
{{{x = 2+sqrt(3)}}} and {{{x = 2-sqrt(3)}}}

Here's what the graph will look like:
{{{graph(300,200,-10,10,-10,10,x^2-4x+1)}}}