1) Faça o estudo do sinal das funções do 1° grau: (y > 0; y < 0) a) f (x) = 2x + 3 b) f (x) = - 3x + 2 c) f (x) = - 5x d) f (x) = 2 x - 3 e) f (x) = -3x + 15 f) f (x) = 2x + 8 g) f(x) = 8x – 16 h) f(x) = 4x + 8 i) f(x) = 2x – 6 j) f(x) = 12x – 4
Soluções para a tarefa
Resposta:
a) f(x) = 2x + 3
a > 0
2x + 3 = 0
2x = - 3
x = -3/2 => raiz da função
f(x) = 0 => x = -3/2
f(x) > 0 => x > -3/2
f(x) < 0 => x < -3/2
b) f(x) = - 3x + 2
a < 0
-3x + 2 = 0
-3x = - 2 . (-1)
3x = 2
x = 2/3
x = 2/3
f(x) = 0 => x = 2/3
f(x) > 0 => x < 2/3
f(x) < 0 => x > 2/3
c) f(x) = - 5x
a < 0
-5x = 0
x = 0
f(x) > 0 => x < 0
f(x) < 0 => x > 0
d) f (x) = 2 x - 3
a > 0
2x - 3 = 0
2x = 3
x = 3/2
f(x) > 0 => x > 3/2
f(x) < 0 => x < 3/2
e) f (x) = -3x + 15
a < 0
-3x + 15 = 0
15 = 3x
x = 5
f(x) > 0 => x < 5
f(x) < 0 => x > 5
f) f (x) = 2x + 8
a > 0
2x + 8 = 0
2x = -8
x = - 4
f(x) > 0 = > x > - 4
f(x) < 0 = > x < - 4
g) f(x) = 8x – 16
a > 0
8x - 16 = 0
8x = 16
x = 2
f(x) > 0 = > x > 2
f(x) < 0 = > x < 2
h) f(x) = 4x + 8
a > 0
4x + 8 = 0
4x = -8
x = -2
f(x) > 0 = > x > - 2
f(x) < 0 = > x < - 2
i) f(x) = 2x – 6
a > 0
2x - 6 = 0
2x = 6
x = 3
f(x) > 0 = > x > 3
f(x) < 0 = > x < 3
j) f(x) = 12x – 4
a > 0
12x - 4 = 0
12x = 4
x = 1/3
f(x) > 0 = > x > 1/3
f(x) < 0 = > x < 1/3