Happy New Year!

Derivate is an option

$y=x^2-4030x+4062240$ $y'=0\Rightarrow 2x-4030=0\\\Rightarrow x=2015$

Now substitute $x=2015$ $y=2015^2-(4030×2015)+4062240\\=Odd-Even+Even\\=Odd$

So out of given choices, $(2015,2015)$ satisfies.

vertex of the parabola is the vertex of the equation: h = -b/2a ; k =( 4ac-b^2)/4a from the G.E. ax^2 + bx + c = 0

$-b\div 2a$ gives the x coordinate of the vertex. The y coordinate of the vertex is just f( $-b\div 2a$ )

vertex of the parabola is the vertex of the equation: h = -b/2a ; k =( 4ac-b^2)/4a from the G.E. ax^2 + bx + c = 0

The vertex of a quadratic equation is $(\frac{-b}{2a},f(\frac{-b}{2a}))$

$\quad y = x^2 -4030x +4062240$

$\quad \Rightarrow y = x^2 -4030x + 4060225 + 2015$

$\quad \Rightarrow y = x^2 -2(2015)x + 2015^2 + 2015$

$\quad \Rightarrow y = (x-2015)^2 + 2015$

Now the vertex here is the lowest point of the equation and that happens when $x-2015)^2$ which is $\ge 0$ is equal to $0$ or when $x = 2015$ and when $x=2015\quad \Rightarrow y = 2015$ . Therefore, the vertex is $(2015,2015)$ .

Happy New Year!

The vertex of the quadratic equation is ( $\frac{-b}{2a}$ , $\frac{-D}{4a}$ ). here,

a=1.

b=-4030.

c=4062240.

D= ${ b }^{ 2 }-4ac$ =4030^2-4 x 1 x 4062240=-8060.(determinant)

Thus, vertex= ( $\frac{4030}{2}$ , $\frac{8060}{4}$ )=(2015,2015)