Question

Sa se determine functia de gradul doi f:R-R f(x)=(3m-1)x²-(2m²+1)x+3, m€R al carui grafic are abscisa vârfului egală cu 3/4

Answer 1

f : IR → IR
f(x) = (3m - 1)x² - (2m² + 1)x + 3, m ∈ IR

$x_v = \frac{3}{4} \\ x_v = \frac{-b}{2a} = \frac{2m^2 + 1}{2(3m - 1)} = \frac{2m^2+1}{6m-2} \\ \frac{3}{4} = \frac{2m^2+1}{6m-2} \\ 8m^2+4 = 18m - 6 \\ 8m^2-18m+10 = 0 | :2 \\ 4m^2-9m+5 = 0 \\ \Delta = 81 - 80 = 1 \\ m_1 = \frac{9+1}{8} = \frac{5}{4} \\ m_2 = \frac{9-1}{8} = 1$

f₁(x) = 11/4 x² - 33/8 x + 3
f₂(x) = 2x² - 3x + 3