Differentiation?

Sol
設切點(a，b)
b=a^3+2a^2－8
y=x^3+2x^2－8
y’=3x^2+4x
m=3a^2+4a
y－b=m(x－a)
y－b=(3a^2+4a)(x－a)
y=(3a^2+4a)x+(b-3a^3－4a^2)
=(3a^2+4a)x+(a^3+2a^2－8－3a^3－4a^2)
=(3a^2+4a)x+(－2a^3－2a^2－8)
So
2a^3+2a^2+8=0
a^3+a^2+4=0
(a^2－a+2)(a+2)=0
(－1)^2－4*1*2=－7<0
a+2=0
a=－2
m=3*4+4*(－2)=4

For graph y=x^3+2x^2-8 : slope, y'=3x^2 + 4x

For line y=mx: slope=m

When line y=mx is tangent to graph y=x^3+2x^2-8 :
i) slope of graph at tangent point = m
=> 3x^2 + 4x = m
=> 3x^2 + 4x - m = 0

ii) line touches graph at only 1 point
=> 3x^2 + 4x - m = 0 has only 1 solution
=> discriminant = 0 *
=> 4^2 - 4(3)(-m) = 0
=> 4(4+3m) = 0
=> 4+3m = 0
=> 3m = -4
=> m = -4/3

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

* For any quadratic eqt. ax^2 + bx - c = o,
x =[ -b ± √(b^2 - 4ac)] /2a , where discriminant = b^2 - 4ac