Question 460473
The 1st step would be to find the axis of symmetry.
This would be a vertical line, {{{ x = x[s] }}}. The parabola
is symmetrical to the left and right of this line.
If the equation is expressed as {{{ ax^2 + b*x + c }}}, then
{{{ x[s] = -b/(2a) }}}. 
The base, or the point with zero slope, will be {{{ x[s], y[s] }}}.
Plug the value of {{{ x[s] }}} back into the equation to get {{{ y[s] }}}
If {{{ a }}} is positive, the parabola opens up.
If {{{ a }}} is negative, it opens down.
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Here's an example:
{{{ y = -2x^2 + 3x - 12 }}}
{{{ a }}} is negative, so it opens down.
{{{ x[s] = (-3)/(2*(-2)) }}}
{{{ x[s] = 3/4 }}} is the axis of symmetry
Plugging this back in:
{{{ y = -2*(3/4)^2 + 3*(3/4) - 12 }}}
{{{ y = -9/8 + 18/8 - 96/8 }}}
{{{ y = -88/8 }}}
{{{ y[s] = -11 }}}
So, the base point is (3/4, -11)
The parabola can touch the x-axis in
1 point, 2 points, or no points.
Here's a plot:
{{{ graph( 400, 400, -5, 5, -25, 5, -2x^2 + 3x - 12 ) }}}
The y-intercept is at {{{ x = 0 }}}, so
{{{ y = -2*(0)^2 + 3*(0) - 12 }}}
{{{ y = -12 }}}, so you have the point (0, -12)