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# Show that the tangents to the curve y = 7x^{3} + 11 at the points where x = 2 and x = - 2 are parallel

**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 slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The equation of the given curve is y = 7x^{3} + 11

dy/dx = 21x^{2}

Thus,

the slope of the tangent at the point where x = 2 is given by,

dy/dx]_{x = 2} = 21 (2)x^{2}

= 84

Also,

the slope of the tangent at the point where x = - 2 is given by,

dy/dx]_{x = - 2} = 21(- 2)^{2}

= 84

It is observed clearly that the slopes of the tangents at the points where x = 2 and x = - 2 are equal.

Hence, the two tangents are parallel

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 16

## Show that the tangents to the curve y = 7x^{3} + 11 at the points where x = 2 and x = - 2 are parallel.

**Summary:**

Since the slopes are equal hence, the tangents to the curve y = 7x^{3} + 11 at the points where x = 2 and x = - 2 are parallel

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