Question

defina k para que f(x)=3x²+x+k+2 tenha o valor mínimo de 31/12

Answer 1

$f(x)=3x^2+x+k+2$

$Y_v=\dfrac{-\Delta}{4a}$

$Y_v=\dfrac{31}{12}$

$\Delta=1^2-4\cdot3\cdot(k+2)=1-12(k+2)=1-12k-24=-12k-23$

$-\Delta=12k+23$

$\dfrac{12k+23}{4\cdot3}=\dfrac{31}{12}$

$\dfrac{12k+23}{12}=\dfrac{31}{12}$

$12k+23=31$

$12k=8$

$k=\dfrac{8}{12}$

$k=\dfrac{2}{3}$

Answer 2

Vamos inicialmente calcular o valor de Δ

$\Delta=1^2-4.3(k+2)=1-12k-24=-12k-23\\ \\ y_V=\frac{-\Delta}{4a}=\frac{12k+23}{12}=\frac{31}{12}\\ \\ 12k+23=31\\ \\ 12k=8\\ \\ k=\frac{8}{12}=\frac{2}{3}$

Agora determinar o valor do mínimo: