Question

Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel.

Answer 1

The equation of the given curve is y = 7x3 + 11.

$\frac{dy}{dx}$= 21x2

The slope of the tangent to a curve at (x0, y0) is $\frac{dy}{dx}$](x0,y0).

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

$\frac{dy}{dx}$]x=-2=21(2)2=84

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

Hence, the two tangents are parallel.