Find the vertex of the equation given.

Equation: y=-2(x-5)(x-1)

the vertex lies on the axis of symmetry.

That axis must lie midway between the roots, because the graph is symmetric.
So, the x-coordinate of the vertex is at (5+1)/2 = 3
y(3) = -2(3-5)(3-1) = -2(-2)(2) = 4
The vertex is at (3,4)

To find the vertex of the equation, we need to determine the coordinates (h, k) of the vertex in the form y = a(x - h)² + k.

In the given equation, y = -2(x - 5)(x - 1), we can rewrite it in the standard form as y = -2(x² - 6x + 5).

In the standard form, the vertex can be found using the following formula:
h = -b/ (2a)
k = f(h)

So, let's find the vertex step by step:

Step 1: Identify the coefficients of the quadratic equation.
a = -2, b = -6, c = 5.

Step 2: Find h.
h = -b / (2a)
h = -(-6) / (2 * -2)
h = 6 / -4
h = -3/2 or -1.5

Step 3: Find k.
To find k, substitute the value of h in the original equation and solve for y.
y = -2(x - 5)(x - 1)
y = -2(-3/2 - 5)(-3/2 - 1)
y = -2(-13/2)(-5/2)
y = -130/4
y = -65/2 or -32.5

Therefore, the vertex of the equation y = -2(x - 5)(x - 1) is (-1.5, -32.5) or written as a pair: (h, k) = (-1.5, -32.5).