Question 404569
The x-coordinate of the vertex of a parabola is at
{{{-b/(2a)}}}, where the equation is in the form:
{{{f(x) = ax^2 + bx + c}}} 
In the given equation:
{{{f(x) = x^2 + 6x - 2}}}
{{{a = 1}}}
{{{b = 6}}}
{{{c =-2}}}
{{{-b/(2z) = -6/(2*1)}}}
{{{-6/(2*1) = -3}}}
Now plug this back into equation to get {{{f(x)}}}
{{{f(-3) = (-3)^2 + 6*(-3) - 2}}}
{{{f(-3) = 9 - 18 - 2}}}
{{{f(-3) = -11}}}
So the vertex is at the point h = (-3,-11)
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I'm not sure what the table is supposed to look like.
It sounds like they want
(-3,-11) (vertex)
(-4, f(-4))
(-5,f(-5))
(-6,f(-6))
etc., and, in the (+) direction from the vertex:
(-2, f(-2))
(-1,f(-1))
(0,f(0))
(1,f(1))
etc.
Here's a plot of the equation. Maybe you can
figure out the table.
{{{ graph( 400, 400, -10, 5, -12, 5, x^2 + 6x - 2) }}}