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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives

Show that the tangents to the curve \(y = 7x^3 + 11\) at the points where \(x = 2\) and \(x = -2\) are parallel.

1 Answer

Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Given : $y=7x^3+11$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=21x^2$
Step 2:
Slope at $x=2$ is
$\large\frac{dy}{dx}_{(x=2)}=$$21(2\times 2)$
$\qquad\quad\;\;=84$
Step 3:
Slope at $x=-2$ is
$\large\frac{dy}{dx}_{(x=-2)}$$=21(-2\times -2)$
$\qquad\quad\;\;=84$
Since the slopes are equal the tangent to the curve at $x=2$ and $x=-2$ are equal.
answered Jul 11, 2013 by sreemathi.v
 

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