# Minimum Value of Quadratic (Tomato subjective 65)

problem: Show that for all real x, the expression $${ax^2}$$ + bx + C ( where a, b, c are real constants with a > 0), has the minimum value  $${\frac{4ac – b^2}{4a}}$$ . Also find the value of x for which this minimum value is attained.

solution: f (x)   $${ax^2}$$ + bx + c

Now minimum  derivative = 0 & 2nd order derivative > 0.

$${\frac{df(x)}{dx}}$$ = 2ax + b
Or $${\frac{d^2f(x)}{dx^2}}$$ =2a
Now 2a> so 2nd order derivative > 0 so $${\frac{d^2f(x)}{dx^2}}$$ = 2.

So minimum occurs when

$${\frac{df(x)}{d(x)}}$$ = 0 or 2ax + b = 0

or 2ax = -b
or x = $${\frac{-b}{2a}}$$ (ans)

At x = $${\frac{-b}{2a}}$$

$${ax^2}$$ + bx + c
= $${a\times {\frac{b^4}{4a^2}}}$$ + $${b\times {\frac{-b}{2a}}}$$ + c
= $${\frac{4ac-b^2}{4a}}$$ (proved)