y(2) = -2^3 = -8
y(2+h) = 2 - (2+h) - (2+h)^3
= 2 - 2 - h -8 -12h -6h^2 -4h^3
= -8 -13 h -6 h^2 -4 h^3
y(2+h) -y(2) = -13 h -6 h^2 -4h^3
divide that by h
[y(2+h)-y(2)]/h = -13 -6h -4h^2
when h --->0 that is
slope = -13 Caramba!
now the equation
y = -13 x + b
include point (2,-8)
-8 = -13(2) + b
b = 26 - 8 = 18
so
y = -13x+18
Given the function:
y = 2 - x - x^3
use limits to find the slope (and then the equation) of the tangent line at x =2.
1 answer
1 answer