Question

Sabendo que os zeros da função quadrática f são x1 = -6 e x2 = 1 e que seu gráfico intersecta o eixo y no ponto de coordenada (0, 2). Qual a lei de formação dessa função?

Answer 1

$\Large\boxed{\begin{array}{l}\sf o\,gr\acute afico\,passa\,pelos\,pontos\\\sf A(-6,0),B(1,0)~e~C(0,2).\\\sf f(x)=ax^2+bx+c\\\sf f(0)=a\cdot 0^2+b\cdot 0+c\\\sf c=2\\\sf f(-6)=a\cdot(-6)^2+b\cdot(-6)+2\\\sf 36a-6b+2=0\\\sf 36a-6b=-2\div2\\\sf 18a-3b=-1\\\sf f (1)=a\cdot1^2+b\cdot1+2\\\sf a+b+2=0\\\sf a+b=-2 \end{array}}$

$\Large\boxed{\begin{array}{l}\begin{cases}\sf a+b=-2\\\sf18a-3b=-1\end{cases}\\\\+\underline{\begin{cases}\sf 3a+\diagdown\!\!\!\!\!3b=-6\\\sf 18a-\diagdown\!\!\!\!\!3b=-1\end{cases}}\\\\\sf 21a=-7\\\sf a=-\dfrac{7}{21}\\\\\sf a=-\dfrac{1}{3}\\\\\sf a+b=-2\\\sf -\dfrac{1}{3}+b=-2\cdot3\\\\\sf -1+3b=-6\\\sf 3b=1-6\\\sf 3b=-5\\\\\sf b=-\dfrac{5}{3}\\\\\sf f(x)=-\dfrac{1}{3}x^2-\dfrac{5}{3}x+2\end{array}}$