Question

Determine os coeficientes a, b, c e d da função polinomial p(x) = ax3 + bx2 + cx + d, cujo gráfico passa

pelos pontos P1 = (0, 10), P2 = (1, 7), P3 = (3, −11) e P4 = (4, −14).​

Answer 1

Vamos substituir no polinômio p(x) os pontos dados pela questão:

$P1:~~10~=~a.(0)^3+b.(0)^2+c.(0)+d\\\\P1:~~10~=~0+0+0+d\\\\P1:~~\boxed{d~=~10}\\\\\\P2:~~7~=~a.(1)^3+b.(1)^2+c.(1)+d\\\\P2:~~7~=~a~.~1+b~.~1+c~.~1+10\\\\P2:~~7~=~a+b+c+10\\\\P2:~~\boxed{a+b+c~=\,-3}\\\\\\P3:~~-11~=~a.(3)^3+b.(3)^2+c.(3)+d\\\\P3:~~-11~=~a~.~27+b~.~9+c~.~3+10\\\\P3:~~27a+9b+3c~=\,-21\\\\P3:~~\boxed{9a+3b+c~=\,-7}\\\\\\P4:~~-14~=~a.(4)^3+b.(4)^2+c.(4)+d\\\\P4:~~-14~=~a~.~64+b~.~16+c~.~4+10\\\\P4:~~64a+16b+4c~=~-24\\\\P4:~~\boxed{16a+4b+c~=~-6}$

Perceba que ficamos com 3 equações e 3 incógnitas (a, b e c), ou seja, temos um sistema de equações. Podemos resolver este sistema por qualquer método conhecido, vou utilizar o escalonamento.

$\left\{\begin{array}{ccc}a+b+c~=\,-3\\9a+3b+c~=\,-7\\16a+4b+c~=~-6\end{array}$

$L2\leftarrow L2-L1\\L3\leftarrow L3-L1\\\\\left\{\begin{array}{ccc}a+b+c~=\,-3\\8a+2b~~~~=\,-4\\15a+3b~~~~=~-3\end{array}$

$Simplificar~L2~por~2\\Simplificar~L3~por~3\\\\\left\{\begin{array}{ccc}a+b+c~=\,-3\\4a+b~~~~=\,-2\\5a+b~~~~=~-1\end{array}$

$L3\leftarrow L3-L2\\\\\left\{\begin{array}{ccc}a+b+c~=\,-3\\4a+b~~~~=\,-2\\~~~~a~~~~~~=~1\end{array}$

Com o sistema devidamente escalonado, podemos determinar o valor das incógnitas.

$\boxed{a~=~1}~~~Direto~do~sistema~escalonado\\\\\\\boxed{4a+b~=\,-2}~~~L2~do~sistema~escalonado\\\\4.(1)+b~=\,-2\\\\b~=\,-2-4\\\\\boxed{b~=~-6}\\\\\\\boxed{a+b+c~=\,-3}~~~L1~do~sistema~escalonado\\\\1+(-6)+c~=\,-3\\\\-5+c~=\,-3\\\\c~=\,-3+5\\\\\boxed{c~=~2}$

Resposta: Os coeficientes a, b, c  e  d valem, respectivamente, 1, -6, 2 e 10.