# Maxima and Minima of a Quadratic Equation

*f*(

*x*) =

*ax*

^{2}+

*bx*+

*c*, if

*a*> 0.

Finding Minima

# Graphical method

At

*x*= −*b*/2*a*,*f*(*x*) will have the minimum value = − D/4a# Equation method

In this method, the objective is to break the equation into a format of (P)^{2}± Q. Now, as we know that any perfect square can have a minimum value of 0, so the minimum value of

*f*(

*x*) = ± Q

Let us assume

*f*(

*x*) =

*x*

^{2}– 5

*x*+ 6

*f*(

*x*) =

Since can have a minimum value of zero, so the minimum value of

*f*(

*x*) here is (– 0.25).

Using the graphical method, at

*x*= −

*b*/2

*a*= 5/2, the minimum value of

*f*(

*x*) = −D/4

*a*= −1/4 = (−0.25)

*Finding maxima*As the value of x is increased, the value of y also increases. So, the maximum value of y is + ∝.

*f*(

*x*) =

*ax*

^{2}+

*bx*+

*c*, if

*a*< 0.

*Finding minima*As the value of x is increased, the value of y decreases (since a < 0). So, the minimum value of y is −∝.

It is obvious from the graph that it is going downward till negative of infinity.

*Finding maxima*The same method of finding minima is applied as in the earlier case (a > 0).