Question

Cine știe??????????Dau bifa….

Answer 1

Răspuns:

Explicație pas cu pas:

a) f(x) = 4x-2

∩(Ox) <=> f(x) = 0 => 4x-2 = 0 => x = 1/2 => A(1/2 ; 0)

∩(Oy) <=> x = 0 => f(0) = -2  => B(0 ; -2)

b) f(x) = -√2x+4

∩(Ox) <=> f(x) = 0 => -√2x+4 = 0 => √2x = 4 => x = 4/√2 = 4√2/2

=> x = 2√2  => A(2√2 ; 0)

∩(Oy) <=> x = 0 => f(0) = 4 => B(0 ; 4)

c) f(x) = x/2 -3

∩(Ox) <=> f(x) = 0 => x/2 - 3 = 0 => x/2 = 3 => x = 6 => A(6 ; 0)

∩(Oy) <=> x = 0 => f(0) = -3 => B(0 ; -3)

d) f(x) = -3

∩(Ox) <=> f(x) ≠ 0 => nu se intersecteaza cu Ox

∩(Oy) <=> x = 0 => f(0) = -3 => B(0 ; -3)

e) f(x) = (1-√2)x - √12 +2

∩(Ox) <=> f(x) = 0 =>  (1-√2)x - √12 +2 = 0 =>

(1-√2)x = √12-2 = 2√3 - 2 => x = (2√3-2)/(1-√2) =>

x = (2√3-2)·(1+√2) : [(1-√2)(1+√2)] =>

x = (2√3+2√6-2-2√2) / (-1) =>

x = 2+2√2-2√3-2√6  => A(  2+2√2-2√3-2√6  ; 0)

∩(Oy) <=> x = 0 => f(0) = 2-√12  => B(0 ; 2-√12)

f) f(x) = -3x-(1-3x) = -3x-1+3x = -1

∩(Ox) <=> f(x) ≠ 0 =>  nu se intersecteaza cu Ox

∩(Oy) <=> x = 0 => f(0) = -1 => B(0 ; -1)

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