od Daniel » Petak, 11. Avgust 2017, 14:29
To je OK, a može se uraditi i tako što bismo razvili ovaj kvadrat binoma na desnoj strani:
[dispmath]ax^2+bx+c=a(x-x_0)^2+y_0\\
\cancel{ax^2}+bx+c=\cancel{ax^2}-2ax_0x+ax_0^2+y_0[/dispmath] Pa onda, izjednačavanjem članova sa [inlmath]x[/inlmath] i slobodnih članova, dobijamo:
[dispmath]bx=-2ax_0x\quad\Longrightarrow\quad x_0=-\frac{b}{2a}\\
c=ax_0^2+y_0\quad\Longrightarrow\quad y_0=c-ax_0^2=c-a\left(-\frac{b}{2a}\right)^2=\cdots=\frac{4ac-b^2}{4a}[/dispmath]
Ako bismo to primenili na tvoju parabolu,
[dispmath]x^2-x=a(x-x_0)^2+y_0=ax^2-2ax_0x+ax_0^2+y_0[/dispmath] odatle se lako dobije, grupisanjem odgovarajučih članova, da je [inlmath]a=1[/inlmath], [inlmath]x_0=\frac{1}{2}[/inlmath] i [inlmath]y_0=-\frac{1}{4}[/inlmath].
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain