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)

⇒ = 3x^{3 – 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