y = (x-1)(x+3)
y = (x - 1)(x - (-3))

when y = 0, x = 1 or x = -3

The two points (-3,0) and (1,0) of the symmetrical quadratic curve are equidistant from each other, and also from the minimum point.

The x-coordinate of the minimum point is the mid point of -3 and 1.

= (1 + (-3)) ÷ 2
= -2/2
= -1

When x = -1, y = (-1 - 1)(-1 + 3)
y = -2 x 2 = -4

So minimum point is (-1,-4)

Alternatively, use differentiation.

y = (x-1)(x+3) = x² +2x - 3

dy/dx = 2x + 2

At minimum point, dy/dx = 0

2x + 2 = 0
2x = -2
x = -1

when x = -1, y = - 4 (see previous comment for this step's working if you're unsure)


So coordinates are (-1,-4)