Question

If the tangent to the curve y=xcube+ax+b at p(1,-6) is parallel to the line y-x=0, find the value of a and b

Answer 1

Hey!

Your answer is in the following attachement.

Answer 2

The values of a and b are -2 and - 5 respectively

• Given curve is

             $y=x^{3}+ax+b$          ----------------------(1)

         Differentiating on both sides w.r.t 'x'

             $\frac{dy}{dx}=3x^{2} +a$

• The slope of tangent to (1) at P(1, -6) is given by

              $m=\frac{dy}{dx} |_{(1, -6)}$

              $m=3(1)^{2} +a$

              $m=3+a$                   -----------------------(2)

• Given that tangent is parallel to the line

                y - x = 0

                y = x

                m = 1

• Substituting m in (2), we get

                 1 = 3 + a

                 a = 1 - 3

                 a = -2

• (1) passes through (1, -6) and a = -2, substituting these values in (1)

                 $-6 = (1)^{3}+(-2)(1)+b$

                 -6 = 1 - 2 + b

                   b = - 6 - 1 + 2

                  b = - 5

       ∴ The value of a is - 2 and b is - 5