What is the range of a quadratic function?
Precalculus Functions Defined and Notation Range
The range of #f(x) = ax^2+bx+c# is:
#{ ([c-b^2/(4a), oo) " if " a > 0), ((-oo, c-b^2/(4a)] " if " a < 0) :}#
Given a quadratic function:
#f(x) = ax^2+bx+c" "# with #a != 0#
We can complete the square to find:
#f(x) = a(x+b/(2a))^2+(c-b^2/(4a))#
For real values of #x# the squared term #(x+b/(2a))^2# is non-negative, taking its minimum value #0# when #x = -b/(2a)#.
#f(-b/(2a)) = c - b^2/(4a)#
If #a > 0# then this is the minimum possible value of #f(x)# and the range of #f(x)# is #[c-b^2/(4a), oo)#
If #a < 0# then this is the maximum possible value of #f(x)# and the range of #f(x)# is #(-oo, c-b^2/(4a)]#
Another way of looking at this is to let #y = f(x)# and see if there's a solution for #x# in terms of #y#.
#y = ax^2+bx+c#
Subtract #y# from both sides to find:
#ax^2+bx+(c-y) = 0#
The discriminant #Delta# of this quadratic equation is:
#Delta = b^2-4a(c-y) = (b^2-4ac)+4ay#
In order to have real solutions, we require #Delta >= 0# and so:
#(b^2-4ac)+4ay >= 0#
Add #4ac-b^2# to both sides to find:
#4ay >= 4ac-b^2#
If #a > 0# then we can simply divide both sides by #4a# to get:
#y >= c-b^2/(4a)#
If #a < 0# then we can divide both sides by #4a# and reverse the inequality to get:
#y <= c-b^2/(4a)#
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