Bonjour
KleinK
f(x) = -6x² + 7x - 2
1) (* est le signe de la multiplication)
(3x - 2)(-2x + 1) = 3x * (-2x) + 3x * 1 - 2 * (-2x) - 2 * 1
(3x - 2)(-2x + 1) = - 6x^2 + 3x + 4x - 2
(3x - 2)(-2x + 1) = - 6x^2 + 7x - 2
(3x - 2)(-2x + 1) = f(x)
2) f(-2) = - 6 * (-2)² + 7 * (-2) - 2
f(-2) = - 6 * 4 - 14 - 2
f(-2) = - 24 - 14 - 2
[tex]\boxed{f(-2) = -40}\\\\f(\dfrac{3}{4})=-6\times(\dfrac{3}{4})^2+7\times\dfrac{3}{4}-2\\\\f(\dfrac{3}{4})=-6\times\dfrac{9}{16}+\dfrac{21}{4}-2\\\\f(\dfrac{3}{4})=-3\times\dfrac{9}{8}+\dfrac{21}{4}-2\\\\f(\dfrac{3}{4})=-\dfrac{27}{8}+\dfrac{42}{8}-\dfrac{16}{8}\\\\\boxed{f(\dfrac{3}{4})=-\dfrac{1}{8}}[/tex]
[tex]\\\\f(-\dfrac{2}{3})=-6\times(-\dfrac{2}{3})^2+7\times(-\dfrac{2}{3})-2\\\\f(-\dfrac{2}{3})=-6\times\dfrac{4}{9}-\dfrac{14}{3}-2\\\\f(-\dfrac{2}{3})=-\dfrac{24}{9}-\dfrac{42}{9}-\dfrac{18}{9}\\\\f(-\dfrac{2}{3})=-\dfrac{84}{9}\\\\\boxed{f(-\dfrac{2}{3})=-\dfrac{28}{3}}[/tex]
3) Le point A(3/4;-0,125) appartient à la courbe car f(3/4) = -1/8 = -0,125 (voir question 2)
Le point B(-2;-40) appartient à la courbe car f(-2) = -40 (voir question 2)
Le point C(-2/3;-9,3) n'appartient pas à la courbe car f(-2/3) = -28/3 = -9,333333... ≠ -9,3 (voir question 2)
Le point E(2/3;0) appartient à la courbe car f(2/3) = 0.
En effet :
[tex]f(\dfrac{2}{3})=(3\times\dfrac{2}{3}-2)(-2\times\dfrac{2}{3}+1)\\\\f(\dfrac{2}{3})=(2-2)(-2\times\dfrac{2}{3}+1)\\\\f(\dfrac{2}{3})=0\times(-2\times\dfrac{2}{3}+1)\\\\\boxed{f(\dfrac{2}{3})=0}[/tex]
