C) f( x) = 4/5 x+ 1/5
1) Gf ∧ Ox ⇒ f( x) = 0
4/5 x + 1/5 = 0 I · 5
4X + 1 = 0
4X = - 1
X = - 1/4 ⇒ f( - 1/4) = 0 ⇒ C( - 1/4, 0)
Gf ∧ Ox = { C{ - 1/4, 0)}
2) Gf ∧ Oy ⇒ x = 0
f ( 0) = 4/5 · 0 + 1/5
f( 0) = 1/5 ⇒ D ( 0 , 1/5)
Gf ∧ Oy = { D { 0, 1/5)}