Question

seja f(x)=ax²+bx+c, sabendo que f(-1)=-5, f(0)=-4, e f(1)=1, qual o valor de f(2)

Answer 1

$\mathbf{Pontos}\ \to\ \mathrm{(-1,-5);\ (0,-4);\ (1,1)}\\\\ \mathbf{Utilizando\ (0,-4),\ obteremos:}\\\\ \mathrm{f(x)=ax^2+bx+c\ \to\ -4=a.0^2+b.0+c\ \to\ \boxed{\mathrm{c=-4}}}\\\\ \mathbf{Utilizando\ (-1,-5),\ obteremos:}\\\\ \mathrm{f(x)=ax^2+bx-4\ \to\ -5=a.(-1)^2+b.(-1)-4\ \to}\\\\ \mathrm{-5=a-b-4\ \to\ a-b=-5+4\ \to\ \boxed{\mathrm{a-b=-1}}\ \mathbf{(I)}}\\\\ \mathbf{Utilizando\ (1,1),\ obteremos:}\\\\ \mathrm{f(x)=ax^2+bx-4\ \to\ 1=a.1^2+b.1-4\ \to\ \boxed{\mathrm{a+b=5}}\ \mathbf{(II)}}$

$\mathbf{Somando\ (I)\ e\ (II),\ obteremos:}\\\\ \mathrm{(a-b)+(a+b)=-1+5\ \to\ 2a=4\ \to\ \boxed{\mathrm{a=2}}}\\\\ \mathbf{Substituindo\ a\ em\ (II),\ obteremos:}\\\\ \mathrm{a+b=5\ \to\ b=5-a\ \to\ b=5-2\ \to\ \boxed{\mathrm{b=3}}}\\\\ \mathbf{Portanto:}\\\\ \mathrm{f(x)=ax^2+bx-4}\ \to\ \boxed{\mathrm{f(x)=2x^2+3x-4}}\\\\ \mathbf{Logo:}\\\\ \mathrm{f(2)=2.2^2+3.2-4=2.4+6-4=8+2=10}\\\\ \boxed{\boxed{\mathbf{f(2)=10}}}$