Asked by David
4. For the curve with y = -2x3 + x - 6, find:
(b.) the gradient at the point (-1, -5)
(c.) the equation of the tangent line at the point (-1, -5).
(b.) the gradient at the point (-1, -5)
(c.) the equation of the tangent line at the point (-1, -5).
Answers
Answered by
Reiny
y = -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
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
Answered by
Hi
y = -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
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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