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Maxima and Minima of a Quadratic Equation

f(x) = ax2 + bx + c, if a > 0.

Finding Minima

Graphical method


 
At x = − b/2af(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) = x2 – 5x + 6
f(x) = Description: 2158.png

Since Description: 2167.png 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/2a = 5/2, the minimum value of f(x) = −D/4a = −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 + ∝.
 
It is obvious from the graph that it is going upwards till infinity.
f(x) = ax2 + 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).





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