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March 6, 2015

March 6, 2015

Posted by **David** on Thursday, April 1, 2010 at 11:03pm.

(b.) the gradient at the point (-1, -5)

(c.) the equation of the tangent line at the point (-1, -5).

- math -
**Reiny**, Thursday, April 1, 2010 at 11:18pmy = -2x^3 + x - 6

dy/dx = -6x + 1

at (-1,-5), dy/dx = -6(-1) + 1 = 7

so y = -7x + b

at (-1,-5)

-5 = -7(-1) + b

b = -12

equation of tangent is y = -7x - 12

- math -
**Hi**, Tuesday, April 6, 2010 at 7:59pmy = -2x^3 + x - 6

dy/dx = -6x + 1

at (-1,-5), dy/dx = -6(-1) + 1 = 7

so y = -7x + b

at (-1,-5)

-5 = -7(-1) + b

b = -12

equation of tangent is y = -7x - 12

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