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

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.