Question 64443
Graph: {{{f(x) = 3x^2-6x-1}}}
Here's the graph:
{{{graph(300,200,-5,5,-5,5,3x^2-6x-1)}}}
The equation of the line of symmetry is given by:
{{{x = -b/2a}}} The a and the b come from the general form of the quadratic equation: {{{ax^2+bx+c=0}}}
In your equation, a = 3 and b = -6, so:
{{{(-b)/2a = (-(-6))/2(3)}}}
{{{x = 1}}} is the equation of the line of symmetry.
Since the line of symmetry passes thrugh the vertex, you can find the coordinates of the vertex by substituting x = 1 into the quadratic equation and solving for y.
{{{y = 3x^2-6x-1}}} Substitute x=1 and solve for y.
{{{y = 3(1^2)-6(1)-1}}} Simplify.
{{{y = -4}}}
The vertex point is at (1, -4) Confirm this on the graph.
The graph shows that the vertex point is a minimum. 
To find four more points to meet your requirement, simply choose four conveniently small values of x and substitute these, one-by-one into the quadratic equation and solve for four values of y.
Let x = 0 for point1.
{{{y = 3(0^2)-6(0)-1}}}
{{{y = -1}}} 
Point1 is (0, -1)
Let x = 1 for point2. You've got this one already, it's the vertex point (1, -4)
Let x = 2 for point3.
{{{y = 3(2^2)-6(2)-1}}}
{{{y = 12-12-1}}}
{{{y = -1}}}
Point3 is (2, -1)
I hesitate to deprive you of the pleasure of doing your own math so perhaps you can finish this yourself.