How do you find the slope of a tangent line to the graph of the function #f(x)=x^3-2x+1# at (-1,2)?

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Answer 1 · 2016-10-23T13:44:20

#y'=1# at #(-1,2)#

First, we have to differentiate the function,
#y=x^3-2x+1#

the rule for derivatives is that if.
#y=x^n#
#dy/dx=nx^(n-1)#

so for,

#y=x^3-2x+1#

#dy/dx=3x^(3-1)-2x^0 +0#

#dy/dx=3x^(2)-2#

so to find the gradient of the tangent we sub in the x value of the point.

#dy/dx=3(-1)^2-2#

#dy/dx=3-2#

#dy/dx=1#

The gradient of the tangent is 1 at the point (-1,2)