If the tangent to the curve y = x3 + ax + b at (1, – 6) is parallel to the line x – y + 5 = 0, find a and b

Given:


The Slope of the tangent to the curve y = x3 + ax + b at


(1, – 6)


First, we will find The Slope of tangent


y = x3 + ax + b


(x3) + (ax) + (b)


⇒ = 3x3 – 1 + ) +


⇒ = 3x2 +


The Slope of the tangent to the curve y = x3 + ax + b at


(1, – 6) is


⇒ = 3(1)2 +


⇒ = 3 + ...(1)


The given line is x – y + 5 = 0


y = x + 5 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.


so the The Slope of the line is y = 1x + 5


so The Slope is 1. ...(2)


Also the point (1, – 6) lie on the tangent, so


x = 1 & y = – 6 satisfies the equation, y = x3 + ax + b


i.e, – 6 = 13 + a + b


⇒ – 6 = 1 + a + b


⇒ a + b = – 7 ...(3)


Since, the tangent is parallel to the line, from (1) & (2)


Hence,


3 + = 1


⇒ a = – 2


From (3)


a + b = – 7


⇒ – 2 + b = – 7


⇒ b = – 5


So the value is a = – 2 & b = – 5


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