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 ....
Determine the slope of the tangent to the curve y=2(x^2+x-1)^3 at (-1, -2)
2 answers
So
y2-y1=m(x2-x1)
y-(-2)=-6(x-(-1)
y=6x-1
therefore the slope of tangent is 6?
y2-y1=m(x2-x1)
y-(-2)=-6(x-(-1)
y=6x-1
therefore the slope of tangent is 6?