Question

Para que a parábola da equação y=ax(ao quadrado)+ bx - 1 contenha os pontos (-2,1) e (3,1), os valores se a e b são? Com a resolução, por favor.

Answer 1

$\begin{cases}y=ax^2+bx-1~~(i)\\ y=ax^2+bx-1=y~~(ii)\end{cases}\begin{cases}a\cdot(-2)^2+b\cdot(-2)-1=1~~(i)\\ a\cdot3^2+b\cdot3-1=1~~(ii)\end{cases}\\\\\\ \begin{cases}4a-2b=1+1~~(i)\\ 9a+3b=1+1~~(ii)\end{cases}\begin{cases}4a-2b=2~~(i)~~(multiplica~a~eq~i,~por~1,5)\\ 9a+3b=2~~(ii)\end{cases}$

$\begin{cases}6a-3b=-3~~(i)\\ 9a+3b=2~~(ii)\end{cases}~~(agora~soma~as~duas)\\\\\\ \begin{cases}6a-3b=-3~~(i)\\ 9a+3b=2~~(ii)\end{cases}\\ ~~~~~------\\ ~~~~15a~~~=~~~-1\\\\ ~~~~~~\Large\boxed{a=- \dfrac{1}{15}}$

Então..

$9a+3b=2\\\\ 9\cdot\left( -\dfrac{1}{15}\right)+3b=2\\\\ - \dfrac{9}{15}+3b=2\\\\ - \dfrac{3}{5}+3b=2\\\\ 3b=2+ \dfrac{3}{5}\\\\ 3b= \dfrac{13}{5}\\\\ \Large\boxed{b= \dfrac{13}{15}}$