At what points on the graph of f(x)=2x^3-6x^2-27x is the slope of the tangent line -9?

(-1, 19) and (3, -81)

Step-by-step explanation:

f(x) = 2x³ − 6x² − 27x

f'(x) = 6x² − 12x − 27

-9 = 6x² − 12x − 27

0 = 6x² − 12x − 18

0 = x² − 2x − 3

0 = (x + 1) (x − 3)

x = -1 or 3

f(-1) = 19

f(3) = -81

The points are (-1, 19) and (3, -81).

answer: h

step-by-step explanation:

step-by-step explanation:

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