I have no idea where to start...i know the slope of the tangent line is the derivative...so do i take the derivative of the function, equate that to -1?? Thanks for any help..
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That's exactly what you do
y = x^3 - 6x + 12
dy/dx = 3x^2 - 6
dy/dx = -1
-1 = 3x^2 - 6
-1 + 6 = 3x^2
5 = 3x^2
5/3 = x^2
x = +/- sqrt(5/3)
x = +/- (1/3) * sqrt(15)
y = x^3 - 6x + 12
dy/dx = 3x^2 - 6
dy/dx = -1
-1 = 3x^2 - 6
-1 + 6 = 3x^2
5 = 3x^2
5/3 = x^2
x = +/- sqrt(5/3)
x = +/- (1/3) * sqrt(15)