Given the curve y = - x ^ 3 + 2x, find the linear equation which passes through point B (2,0) and is tangent to curve C? Derivative = -= I set a point (x0, Y0), and then f (x0 + △ x) - f (x0) / △ x, want to find its derivative function = = but it's so strange, is it - △ x ^ 2-3x0 ^ 2-3 △ xX0 + 1, is it my method problem or wrong calculation? = = because I feel that △ x should all be reduced

Given the curve y = - x ^ 3 + 2x, find the linear equation which passes through point B (2,0) and is tangent to curve C? Derivative = -= I set a point (x0, Y0), and then f (x0 + △ x) - f (x0) / △ x, want to find its derivative function = = but it's so strange, is it - △ x ^ 2-3x0 ^ 2-3 △ xX0 + 1, is it my method problem or wrong calculation? = = because I feel that △ x should all be reduced

Let P (x0, Y0) be the tangent point,
y'=-3x^2+2
k=y'|(x=x0)=-3x0^2+2
k=(y0)/(x0-2)
Y0 = - x0 ^ 3 + 2x0
-3x0^2+2=(-x0^3+2x0)/(x0-2)
-3x0^3+2x0+6x0^2-4=-x0^3+2x0
2x0^3-6x0^2+4=0
x0^3-3x0^2+2=0
x0^2(x0-1)-2(x0^2-1)=0
(x0-1)(x0^2-2x0+2)=0 x0^2-2x0+2>0
x0=1
Tangent point P (1,1)
k=-1
Tangent line equation
y=-x+2