Parabeltangent opgave???

Find f'(x)

find f(2) og f'(2)

y=f'(2)(x-2)+f(2)

Og så det samme for koordinatet 0??

          tangentligning:
                                    y = f '(x_o)(x-x_o) + f(x_o)

          tangent_A:
                                    y = f '(0)(x-0) + f(0)

                                    y = -3x + (-2)

                                    y = -3x - 2

          tangent_B:
                                    y = f '(2)(x-2) + 0

                                    y = 5x - 10

   spidse vinkel mellem linjerne

                                             y = -3x - 2  og   y = 5x - 10

                                        
                                                 |-3 - 5|         8
                        tan(V_spids) =  ------------ =  ---  =  (4/7)
                                               |1+(-3)·5|     14

                       V_spids = tan^-1(4/7) = 29,74º

                       V_stump = 180º - 29,74º = 150,26º

alternativt:

             y = -3x - 2 's hældningsvinkel

             tan(φ₁) = -3

             φ₁ = tan^-1(-3) = -71,57º

             y = 5x - 10 's hældningsvinkel

             tan(φ₂) = 5

             φ₂ = tan^-1(5) = 78,69º

     den stumpe vinkel mellem   y = -3x - 2  og  y = 5x - 10
     er

             V_stump = φ₂ - φ₁ = 78,69º - (-71,57º) = 150,26º

alternativt_2:

                   y = -3x - 2 's retningsvektor er a =[1,-3]      |a| = √(1²+(-3)²) = √(10)

                   y = 5x - 10 's retningsvektor er b =[1,5]      |b| = √(1²+5²) = √(26)

                                             a•b = 1·1 + (-3)·5 = 1 - 15 = -14

                                             |a|·|b| = √(260) = 2√(65)

       vinklen mellem linjerne er lig med vinklen mellem deres retningsvektorer

                                                a•b         -14            -7
                              cos(V) = --------  =  --------  =  -----
                                              |a|·|b|      2√(65)     √(65)

                              V = cos^-1(-7/√(65)) = 150,26º

alternativt_3:

                   3x + y + 2 = 0 's normalvektor er n₁ = [3,1]        |n₁| = √(3²+1²) = √(10)

                   -5x + y - 10 = 0 's normalvektor er n₂ = [-5,1]      |n₂| = √((-5)²+1²) = √(26)

                                            n₁•n₂ = 3·(-5) + 1·1 = +15 + 1 = -14

                                             |n₁|·|n₂| = √(260) = 2√(65)

       vinklen mellem linjerne er lig med vinklen mellem deres normalvektorer

                                              n₁•n₂         -14            -7
                              cos(V) = ---------  =  --------  =  ------
                                              |n₁|·|n₂|     2√(65)     √(65)

                              V = cos^-1(-7/√(65)) = 150,26º

rettelse
                     n₁•n₂ = 3·(-5) + 1·1 = -15 + 1 = -14