How do you find the vertex of a parabola #y = -2(x+1)(x-5)#?

Consider a parabola represented by a quadratic polynomial
#y=ax^2+bx+c#
If #a>0# the parabola's horns are directed upwards and it reaches its minimum at some central point #x=x_0#.
If #a<0# the parabola's horns are directed downwards and it reaches its maximum at some central point #x=x_0#.
In both cases the X-coordinate of a vertex is #x_0# and its Y-coordinate is #ax_0^2+bx_0+c#.

Parabola is a symmetrical figure, its vertical axis of symmetry intersects the X-axis in exactly the same point where it reaches its minimum (for #a>0#) or maximum (for #a<0#) value, that is at point #x=x_0#.

If we are dealing with a quadratic polynomial that can be equal to zero at some values of #x#, that is if the equation
#ax^2+bx+c=0#
has solutions #x=x_1# and #x=x_2#,
the central point #x_0# lies exactly in the middle between these solutions #x_1# and #x_2#.

Therefore, in case we have two solutions mentioned above, all we have to do to find a vertex is
(a) find these two solutions #x_1# and #x_2# of an equation:
#ax^2+bx+c=0#
(b) find a midpoint between them:
#x_0=(x_1+x_2)/2#.

The given problem presents an easy case, when the solutions to an equation #-2(x+1)(x-5)=0# are, obviously, #x_1=-1# and #x_2=5#.
Hence, the midpoint between them is
#x_0=(-1+5)/2=2#

So, we have determined that X-coordinate of the parabola's axis of symmetry as #x_0=2#.
The Y-coordinate is #y_0_ = -2(2+1)(2-5)=18#.
So the vertex is at point #(2,18)#