# if f(x)= ax2 + bx +c and a>0 then prove that the minimumvalue of f(x)= -D/4a at x= -b/2a

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`f(x) = ax^2+bx+c`

When f(x) has a minimum or maximum then `f'(x) = 0` .

`f'(x) = 2ax+b`

When `f'(x) = 0` ;

`2ax+b = 0`

`x = -b/(2a)`

If f(x) has a minimum at `x = -b/(2a)` then `f''(x)>0` .

`f'(x) = 2ax+b`

`f''(x) = 2a`

It is given that `a>0` . So `2a>0` and thus `f''(x)>0` .

So we have a minimum for `f(x)` at `x = -b/(2a)`

Minimum f(x)

`= a(-b/2a)^2+b(-b/2a)+c`

`= b^2/4a-b^2/2a+c`

`= (-b^2+4ac)/(4a)`

Since `b^2-4ac = D` ;

Minimum f(x)

`= (-b^2+4ac)/(4a)`

`= -D/(4a)`

** So f(x) has its minimum of **`-D/(4a) ` at `x = -b/(2a)`

We have given quadratic polynomial

`f(x)=ax^2+bx+c ,a>0`

We can write

`f(x)=a(x^2+(b/a)x+(b/(2a))^2-(b/(2a))^2+c/a)`

`f(x)+a((b/(2a))^2-c/a)=a(x+b/(2a))^2`

`f(x)+(b^2-4ac)/(4a)=a(x+b/(2a))^2`

`` Let define

`D=(b^2-4ac)`

`f(x)+D/(4a)=a(x+b/(2a))^2`

a>0 ,`(x+b/(2a))^2 >=0 ==>f(x)+D/(4a)>=0`

`f(x)+D/(4a)=0 ==> f(x)=-D/(4a)` then `(x+b/(2a))^2=0`

`x=-b/(2a)`

**f(x) has min value -D/(4a) when x=-b/(2a)**

Let a=1>0 ,b=2 and c=1

its graph , f(x) represented along y axis and x along x axis.

so minimum value of f(x)=0 when x=-1