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# Find the points at which tangent to the curve y = x^{3} - 3x^{2} - 9x + 7 is parallel to the x-axis

**Solution:**

For a curve y = f(x) containing the point (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by

y − y_{1} = f′(x_{1}) (x − x_{1}).

The given curve is

y = x^{3} - 3x^{2} - 9x + 7

Therefore,

dy/dx =

d/dx (x^{3} - 3x^{2} - 9x + 7)

= 3x^{2} - 6x - 9

Since tangent is parallel to the x-axis if the slope of the tangent is zero.

Hence,

3x^{2} - 6x - 9 = 0

⇒ x^{2} - 2x - 3 = 0

⇒ (x - 3)(x + 1) = 0

⇒ x = 3 or x = - 1

When, x = 3

Then,

y = (3)^{3} - 3(3)^{2} - 9 (3) + 7

= 27 - 27 - 27 + 7

= - 20

When, x = 1

Then,

y = (- 1)3 - 3(- 1)^{2} - 9(- 1) + 7

= - 1 - 3 + 9 + 7

= 12

Thus, the points at which the tangent is parallel to the x-axis are (3, - 20) and (- 1, 12)

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 7

## Find the points at which tangent to the curve y = x^{3} - 3x^{2} - 9x + 7 is parallel to the x-axis

**Summary:**

The points at which tangent to the curve y = x^{3} - 3x^{2} - 9x + 7 is parallel to the x-axis are (3, - 20) and (- 1, 12). For a curve y = f(x) containing the point (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by y − y_{1} = f′(x_{1}) (x − x_{1})

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