Using given three points #P(3,0), Q(7,0) and R(9,-24)#, let us
first determine the eqn. of the Parabola #S#.
Suppose that, #S : y=ax^2+bx+c, a!=0#.
#P in S rArr 0=a*3^2+b*3+c, i.e., 9a+3b+c=0.........(1)#.
#Q in S rArr 49a+7b+c=0................................................(2)#.
#R in S rArr 81a+9b+c=-24..........................................(3)#.
#(2)-(1)rArr40a+4b=0rArr10a+b=0........................(4)#.
#(3)-(2)rArr32a+2b=-24rArr16a+b=-12............(5)#.
#(5)-(4)rArr6a=-12rArra=-2................................(6)#.
#(6) & (4)rArr b=20............................................................(7)#.
#(6), (7) & (1)rArrc=-42.................................................(8)#.
From #(6), (7) and (8)#, we have,
# S : y=-2x^2+20x-42#
#rArr y+42=-2x^2+20x=-2(x^2-10x)#,
#=-2(ul(x^2-10x+25)-25)#,
#=-2(x^2-10x+25)+50=-2(x-5)^2+50#.
#rArr S : y+42-50=-2(x-5)^2, or, #
# S : y-8=-2(x-5)^2#, from which we conclude that,
#(5,8)# is the desired vertex!
