Revisiting Polynomials

Minimum value of a quadratic equation (whose $a> 0$ ) occurs at

$\boxed {x= \dfrac {-b}{2a}}$

and is equal to :

$\color{#3D99F6}{\boxed{f_{min}= \dfrac {4ac-b^2}{4a}}}$

Substituting the given values we get:

$f_{min}= \dfrac {-1}{12}$

$= \color{#20A900}{\boxed {-0.083}}$

$p(x)=3\left(x-\dfrac{5}{6}\right)^2+\left(-\dfrac{1}{12}\right)$

(See Completing the squares )

Since square of a real quantity is always non negative. $\therefore\left(p(x)\right)_{\text{min}}=\dfrac{-1}{12}\approx-0.0833@x=\dfrac 56$

To calculate the minimum value first we would have to find the point at which its derivative is 0.

Derivative is $\frac { d({ 3x }^{ 2 }-5x+2) }{ dx } =6x-5$ . Then $6x-5=0$ . This occurs at point $x=\frac { -5 }{ 6 }$

Hence then just plug the value in the polynomial p(x). You would get $\frac { -1 }{ 12 }$ which is -0.083.

Let us consider the given equation as $y=3x^2-5x+2$ . $\frac{dy}{dx}=\frac{d}{dx} (3x^2-5x+2)=6x-5$ $If \space 6x-5=0 \space Then \space x=\frac{5}{6}$ Substitute $x=\frac{5}{6}$ in the given equation $=> y=3\left(\frac{5}{6}\right)^2-5\left(\frac{5}{6}\right)+2=\frac{-1}{12}=-0.083$ $\frac{d^2}{dx^2}\left(3x^2-5x+2\right)=6>0$ Therefore Y is minimum at x = $\frac{5}{6}$ Therefore the minima of the following Equation is $\boxed{-0.83}$