In the general case #f(x)=ax^2+bx+c# we can complete the square as follows:
#f(x) = ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))#
The vertex of this parabola is where #x+b/(2a) = 0#, that is when #x = -b/(2a)# and #y = c - b^2/(4a)#.
That is #(-b/(2a), c - b^2/(4a))#
The axis of symmetry is the vertical line given by the equation
#x = -b/(2a)#
The intercept with the #y# axis is #(0, c)#
The intercepts with the #x# axis are the roots of #f(x) = 0# which can be calculated in general as:
#((-b+-sqrt(b^2-4ac))/(2a), 0)#
If #c = 0# this simplifies to:
#((-b+-b)/(2a), 0)#
That is #(-b/a, 0)# and #(0, 0)#
In the case of #f(x)=x^2-5x#, we have #a=1#, #b=-5# and #c=0#.
The vertex is at #(-b/(2a), c-b^2/(4a)) = (5/2, -25/4)#
The axis of symmetry is #x = -b/(2a) = 5/2#
The intercept with the #y# axis is #(0, c) = (0, 0)#
Because #c=0#, the intercepts with the #x# axis are
#(-b/a, 0) = (5, 0)# and #(0, 0)#
