Pembuktian Rumus Diskriminan

by Muhammad Rahmi on May 13, 2014 in KAPSEL SMP

Rumus Diskriminan

Pembuktian :

y=ax^2+bx+c

$=a(x+\frac{bx}{a}+...)+c$
$=a(x+\frac{bx}{a}+(\frac{b}{2a})^2)+c-\frac{b^2}{4a}$
$=a(x+\frac{b}{2a})^2+\frac{4ac}{4a}-\frac{b^2}{4a}$

$=a(x+\frac{b}{2a})^2+(\frac{b^2-4ac}{4a})$

maka

$a> 0(x+\frac{b}{2a})^2\geq 0$
$a< 0(x+\frac{b}{2a})^2\leq 0$

$Ymax=\frac{b^2-4ac}{4a}$ jika a < 0
$Ymin=\frac{b^2-4ac}{4a}$ Jika a > 0

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Pembuktian Rumus Diskriminan

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