```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

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