область значень y=x2-5x

$y=x^2-5x, \\ x^2-2\cdot\frac{5}{2}x+(\frac{5}{2})^2-(\frac{5}{2})^2=y, \\ (x-\frac{5}{2})^2=y+\frac{25}{4}, \\ x-\frac{5}{2}=\sqrt{y+\frac{25}{4}}, \\ x=\sqrt{y+\frac{25}{4}}+\frac{5}{2}, \\ y+\frac{25}{4}\geq0, \\ y\geq-\frac{25}{4}, \\ E_y=[-\frac{25}{4};+\infty).$

II.
$y=x^2-5x, \\ a=1\ \textgreater \ 0, \ b=-5, \ c=0, \\ x_0=-\frac{b}{2a}=-\frac{-5}{2\cdot1}=\frac{5}{2}, \\ y_0=x_0^2-5x_0=(\frac{5}{2})^2-5\cdot\frac{5}{2}=\frac{25}{4}-\frac{25}{2}=-\frac{25}{4}, \\ \{y_0=-\frac{D}{4a}=\frac{4ac-b^2}{4a}=\frac{4\cdot1\cdot0-5^2}{4\cdot1}=\frac{0-25}{4}=-\frac{25}{4}\} \\ y\geq y_0, \ y\geq-\frac{25}{4}.$