Question

Exercițiul 6. Răspuns x+1

Answer 1

Răspuns:

Explicație pas cu pas:

f(x) = mx^2 + 2(m + 1)x + m + 2

Δ = 4(m + 1)^2 - 4m(m + 2) = 4(m^2 + 2m + 1) - 4m^2 - 8m

= 4m^2 + 8m + 4 - 4m^2 - 8m = 4

varful parabolei are coordonatele

x = -2(m + 1)/2m = -(m + 1)/m = -m/m - 1/m = -1 - 1/m

y = -4/4m = -1/m

verificam daca x si y se afla pe dreapta y = x + 1

-1/m = -1 - 1/m + 1

-1/m = -1/m adevarat

Answer 2

$\it V\hat arful\ parabolei\ este\ punctul\ variabil\ \ V(x_V,\ \ y_V),\ \ unde:\\ \\ x_V=-\dfrac{b}{2a}=-\dfrac{2(m+1)}{2m}=-\dfrac{m+1}{m}=-1-\dfrac{1}{m} \Rightarrow x_V+1=-\dfrac{1}{m}\ \ \ \ (1)\\ \\ \\ y_V=-\dfrac{\Delta}{4a}=-\dfrac{4(m+1)^2-4m(m+2)}{4m}=-\dfrac{(m+1)^2-m(m+2)}{4m}=\\ \\ \\ =-\dfrac{m^2+2m+1-m^2-2m}{m}=-\dfrac{1}{m}\ \ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow\ y_V=x_V+1 \Rightarrow\ V(x_V,\ \ y_V)\in\ d,\ \ unde\ \ (d):\ \ y=x+1$