Respuesta :

Edufirst
1) tangent line: slope = f '(2) = 5, point (2,3)

y - 3 = 5(x-2)
y = 5x - 10 + 3
y = 5x - 7

2) normal line: slope = - 1/ f '(2) = - 1/5, point (2,3)

y - 3 = [-1/5](x - 2)

y = - x/5 + 2/5 + 3

y = - x/5 + 17/5