How to compute the minimum and maximum function values of a quadratic function?

Answer 1

Suppose you have a quadratic function of the form

y = ax2 + bx + c where a, b and c are real numbers and a is non-zero. [If a = 0 it is not a quadratic!]

The turning point for this function may be obtained by differentiating the equation with respect to x, or by completing the squares. However you get there, the turning point is the solution to 2ax + b = 0 or x = -b/2a

Now,

if a > 0 then the quadratic has a minimum at x = -b/2a and it has no maximum because y tends to +∞ as x tends to ±∞ .

if a < 0 then the quadratic has a maximum at x = -b/2a and it has no minimum because y tends to -∞ as x tends to ±∞.

You evaluate the value of y at this point.

y = a(-b/2a)2 + b(-b/2a) + c

= b2/4a - b2/2a + c

= -b2/4a + c

= -(b2 - 4ac)/4a

In either case, if the domain of the function is bounded on both sides, then the missing extremum will be at one or the other bound - whichever is further away from (-b/2a).