Asked by Jessi

Determine the slope of the tangent to the curve y=2(x^2+x-1)^3 at (-1, -2)
Answered by Reiny
dy/dx= 6(x^2+x-1)^2 (2x+1)
at (-1,-2)

dy/dx= 6(1-1-1)^2 (-2+1)
= -6

so now you have the slope of -6 and a point(-1,-2)

continue ....
Answered by Jessi
So

y2-y1=m(x2-x1)

y-(-2)=-6(x-(-1)
y=6x-1

therefore the slope of tangent is 6?
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