Please help me with this

1. tangent slope is horizontal at (-5,197) and (1,-19)

2. the tangent line has slope -48 at points (-1, 37) and (-3, 141)

Step-by-step explanation:

1. f(x) = 2x³+12x²-30x-3

f'(x) = 6x²+24x-30

the tangent is horizontal when f'(x) = 0

0 = 6x²+24x-30

0 = x²+4x-5

0 = (x+5)(x-1)

x= -5, 1

f(-5) = -250+300+150-3 = 197

f(1) = -19

tangent slope is horizontal at (-5,197) and (1,-19)

2. when f'(x) = -48

6x²+24x-30 = -48

6x²+24x+18 = 0

x²+4x+3 = 0

(x+1)(x+3) = 0

x = -1, -3

f(-1) = 37

f(-3)=141

the tangent line has slope -48 at points (-1, 37) and (-3, 141)