# f(x)=y=ax^2+bx+c,# intersects the #Y-# Axis in the point
#(0,-5).#
#:. f(0)=-5 rArr c=-5...................(1).#
We know that, for #f_min, f'(x)=0, and, f''(x) gt 0.#
Now, #f(x)=ax^2+bx+c rArr f'(x)=2ax+b, and, f''(x)=2a.#
Since, #f''(x) gt 0, 2a >0, or, a >0.#
#:. f'(x)=0 rArr 2ax+b=0 rArr x=-b/(2a).#
#:. f_min=f(-b/(2a))=a(-b/(2a))^2+b(-b/(2a))+c,#
#=b^2/(4a)-b^2/(2a)+c=c-b^2/(4a)=-5-b^2/(4a).#
This means that #(-b/(2a), -5-b^2/(4a))# is the minimum point of #f.#
#:. (-b/(2a), -5-b^2/(4a))=(2,-13).#
#:. -b/(2a)=2...(2), &, -5-b^2/(4a)=-13 rArr -b^2/(4a)=-8,#
#i.e., b/2(-b/(2a))=-8 :. (2) rArr b/2(2)=-8, or, b=-8.#
Then, by #(2), -(-8)/(2a)=2 rArr a=2; &, a>0," as reqd."#
Altogether, with #a=2, b=-8, and, c=-5,# we get,
#f(x)=2x^2-8x-5,# as desired.
Enjoy Maths.!