Question 1079111
The first two are the vertex of a parabola, and they are minima.
The first has a vertex at x=-1/4 and y=51/8
{{{graph(300,300,-10,10,-10,10,2x^2-x+6)}}}
The vertex is at x=2, y=-1
{{{graph(300,300,-10,10,-10,10,x^2-4x+3)}}}
This has a derivative of 3x^2-3, so critical points are when 3x^2-3=0, which occurs when x=1 or -1
When x=-1, y=5 and the second derivative is 6x and negative when x=-1, so that is a local maximum.  When x=1, a local minimum, y=1.
{{{graph(300,300,-10,10,-10,10,x^3-3x+3)}}}
This has a derivative of -3x^2-12x-9, and setting that equal to 0 and factoring out a -3, x^2+4x+3=0, with stationary points at x=-1,-3 and y=5, 1 or (-1, 5) and (-3, 1).  Second derivative is -6x-12, so when x=-1 second derivative is negative and a local maximum, and when x=-3, it is positive and a local minimum.
{{{graph(300,300,-10,10,-10,10,1-9x-6x^2-x^3)}}}