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f(x) = (1/2)x^2-x-(3/2)        where a = 1/2,  b = -1   and c = -3/2

The x coordinate of the vertex =   -b/ [2a] =   1/[2(1/2)] = 1/1 = 1

And we can substitute back into f(x) to find the y value when x = 1 

So, y = (1/2(1)^2  -(1) -(3/2)   = (1/2) - (5/2)  = -2

So the vertex is at (1. -2)

To get the zeros, set f(x) to 0 and we have

(1/2)x^2-x-(3/2)   = 0    multiply through by 2

x^2 - 2x - 3 = 0     factor

(x - 3) (x + 1)  = 0    and setting each factor to 0 we have that  x = 3 and x = -1

In other words, the zeros occur at (3, 0)  and ( -1, 0)

Here's the graph.......https://www.desmos.com/calculator/syjatyyvrk