For a quadratic function #y=ax^2+bx+c#, a maximum is there if #a<0# and it has a minimum, if #a>0#. Please see below for details.
We can write #y=ax^2+bx+c# as
#y=a(x^2+b/ax)+c#
= #a(x^2+2xxb/(2a)xx x+(b/(2a))^2-(b/(2a))^2)+c#
= #a(x^2+2xxb/(2a)xx x+(b/(2a))^2)-a(b/(2a))^2+c#
= #a(x+b/(2a))^2-b^2/(4a)+c#
= #a(x+b/(2a))^2-(b^2-4ac)/(4a)#
Observe that as #(x+b/(2a))^2# is always greater than #0#,
if #a# is positive, we will have a minima for #y#, when #x+b/(2a)=0# i.e. #x=-b/(2a)#, which will be at #-(b^2-4ac)/(4a)#, and
if #a# is negative, we will have a maxima for #y#, when #x+b/(2a)=0# i.e. #x=-b/(2a)#, which will be at #-(b^2-4ac)/(4a)#.
Hence to find a maxima or minima for a quadratic function, observe the sign of #a# and convert the equation, as above, in form #a(x-h)^2+k#. Then the corresponding maxima or minima will be #k#, when #x=h#.
