1)Касательная проведённая к графику функции y=2x³+12x²+13x-20 в некоторой точке,...

Решите задачу:

$\displaystyle\\f(x)=2x^3+12x^2+13x-20\ \ \ g(x)=-5x+1\\\\\\f'(x)=6x^2+24x+13\ \ \ g'(x)=-5\\\\\\f'(x)=k_1\ \ \ g'(x)=k_2\\\\k_1=k_2\\\\6x^2+24x+13=-5\\\\6x^2+24x+18=0\mid\div6\\\\x^2+4x+3=0\\\\x_1=-3\\x_2=-1\\\\f(-3)=2*(-3)^3+12*(-3)^2+13*(-3)-20=-5\\\\f(-1)=2*(-1)^3+12*(-1)^3+13*(-1)-20=-23\\\\(-3;-5)\ \ \ (-1;-23)\\\\y=f'(x_0)(x-x_0)+f(x_0)\\\\f(x_0)=f(-3)=-5\\\\f'(x_0)=f'(-3)=6*(-3)^2+24*(-3)+13=-5\\\\y_1=-5(x+3)-5=-5x-15-5=-5x-20\\\\f(x_0)=f(-1)=-23\\\\f'(x_0)=6*(-1)^2+24*(-1)+13=-5\\$

$y_2=-5(x+1)-23=-5x-5-23=-5x-28$

$\displaystyle\\2)\ f(x)=-\frac{x^4}{27} +\frac{x^2}{8} -2x+5\ \ \ \ x_0=3\\\\y=f'(x_0)(x-x_0)+f(x_0)\\\\f(x_0)=-\frac{3^4}{27}+\frac{3^2}{8}-2*3+5=-\frac{23}{8}\\\\\\ f'(x)=-\frac{4}{27}x^3+\frac{1}{4}x-2\\\\\\ f'(x_0)=-\frac{4}{27}*3^3+\frac{1}{4}*3-2=-\frac{21}{4}\\\\\\ y=-\frac{21}{4}(x-3)-\frac{23}{8}$