Question

At what points will be tangents to the curve y = 2x³ – 15x² + 36x – 21 be parallel to the x – axis? Also, find the equations of the tangents to the curve at these points.

Answer 1

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=6x^2-30x+36\)

According to the question, tangent is parallel to the x – axis , which implies m = 0

6x² - 30x + 36 = 0

x² - 5x + 6 = 0

x = 3 or x = 2

since this point lies on the curve, we can find y by substituting x

y = 2(3)³ – 15(3)² + 36(3) – 21

y = 6

or

y = 2(2)³ – 15(2)² + 36(2) – 21

y = 7

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 6 = 0(x – 3)

y = 6

or

y – 7 = 0(x – 2)

y = 7