Question 214893
{{{y = -(x+4)(x-2)}}}
{{{y = -(x^2 + 2x - 8)}}}
{{{y = -x^2 - 2x + 8}}}
If the coefficient of the squared term is 
negative ({{{-1}}} in this case), then
the parbola opens down.
The axis of symmetry is at {{{x = -b/(2a)}}}
when the general form of the equation is
{{{ax^2 + 2b + c = 0}}}, so
{{{a = -1}}}
{{{b = -2}}}
{{{x = -(-2)/(2*(-1))}}}
{{{x = 2/(-2)}}}
The axis 0f symmetry is {{{x = -1}}}
and, when {{{x = -1}}},
{{{y = -(x+4)(x-2)}}}
{{{y = -(-1+4)(-1-2)}}}
{{{y = -(3*(-3))}}}
{{{y = 9}}}
The vertex is at (-1,9)
also
{{{y = -(x+4)(x-2)}}}
{{{0 = -(-4 + 4)(-4-2)}}}
There is a zero at {{{x = -4}}}
{{{0 = -(2 + 4)(2-2)}}}
There is a zero at {{{x = 2}}}
{{{ graph( 600, 600, -5, 5, -3, 12,-x^2 - 2x + 8) }}}