Question

A parábola representada pela equação y= ax² + bx passa pelos pontos A= (1/3,1) e B= (1,2). Calcule os valores de a e b

Answer 1

Ae Mikael,

vamos identificar alguns dados:

$\begin{cases}\mathsf{x= \dfrac{1}{3}~~e~~1 }\\\\\mathsf{y=1~~e~~2}\end{cases}$

vamos então montar um sistema de equações do 1° grau, nas incógnitas (a) e (b), tomando a função do 2° grau, vejamos:

$\begin{cases}\mathsf{y=ax^2+bx~~(i)}\\ \mathsf{y=ax^2+bx~~(ii)}\end{cases}\\\\\\ \mathsf{Substituindo,~teremos}\\\\\\ \begin{cases}\mathsf{a\cdot \dfrac{1}{3} ^2+b\cdot \dfrac{1}{3}=1~~(i) }\\\\\mathsf{a\cdot1^2+b\cdot1=2~~(ii)}\end{cases}\\\\\\ \begin{cases}\mathsf{ \dfrac{1}{9}a+ \dfrac{1}{3}b=1~~(i) }\\\mathsf{a+b=2~~(ii)}\end{cases}$

$\mathsf{a=2-b~~(ii)}\\\\ \mathsf{ \dfrac{1}{9}\cdot(2-b)+ \dfrac{1}{3}b=1~~(ii~em~~i) }\\\\ \mathsf{ \dfrac{2-b}{9}+ \dfrac{b}{3} =1 }\\\\ \mathsf{ \dfrac{1\cdot(2-b)+3\cdot b}{\not9}= \dfrac{3\cdot3}{\not9}}\\\\ \mathsf{2-b+3b=9}\\ \mathsf{2b=7}\\\\ \boxed{\mathsf{b= \dfrac{7}{2} }}$

Achamos (b), agora acharemos (a):

$\mathsf{a+b=2}\\\\ \mathsf{a+ \dfrac{7}{2} =2}\\\\ \mathsf{a=2- \dfrac{7}{2} }\\\\ \boxed{\mathsf{a=- \dfrac{3}{2} }}$

Pronto, agora vamos escrever a função do 2° grau:

$\large\boxed{\mathsf{y=- \dfrac{3}{2}x^2+ \dfrac{7}{2}x} }$

Tenha ótimos estudos ;P