Question

Find the point on the curve y = x³ − 11x + 5 at which the tangent is y = x − 11.

 

Answer 1

The equation of the given curve is y = x³ − 11x + 5.

The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c).

∴Slope of the tangent = 1

When x = 2, y = (2)³ − 11 (2) + 5 = 8 − 22 + 5 = −9.

When x = −2, y = (−2)³ − 11 (−2) + 5 = −8 + 22 + 5 = 19.

Hence, the required points are (2, −9) and (−2, 19).
But, both these points should satisfy the equation of the tangent as there would be point of contact between tangent and the curve.
∴ (2, −9) is the required point as (−2, 19) is not satisfying the given equation of tangent.