Question

напишите уравнение касательной к графику
у=3х^2+4х-2; х0=-3

Answer 1

$y = 3x^2 + 4x - 2 \\ y' = 6x + 4 \\ y'(-3) = -18 + 4 = -14 \\ y_k = f(x_0)+f'(x_0)(x - x_0) \\ f(-3) = 27 - 12 - 2 = 13 \\ y_k = 13 - 14(x+3) \\ y_k = 13 - 14x - 42 \\ y_k = -14x -29$

Answer 2

$y=3 x^{2} +4x-2\\x _{0} =-3\\\\y'=f'( x_{0} )(x- x_{0} )+f( x_{0} ) \\ f'(x)=6x+4\\f'( x_{0} )=6*(-3)+4=-18+4=-14\\f( x_{0} )=3*(-3) ^{2} +4*(-3)-2=3*9-12-2= \\ =27-12-2=15-2=13\\y'=-14*(x+3)+3=-14x-43+13=-14x-29$