In this problem
we have
y=x^2-x-8
this is a vertical parabola open upward
the vertex represents a minimum
Convert the given equation into vertex form
so
y=a(x-h)^2+k
where
(h,k) is the vertex of the parabola
y=x^2-x-8
Complete the squares
y=(x^2-x+1/4)-8-1/4
rewrite as perfect squares
y=(x-1/2)^2-33/4
y=(x-0.5)^2-8.25
the vertex is (0.5,-8.25)
the minimum value is the y-coordinate of the vertex
therefore
the answer is -8.25