Show that the tangents to the curve y = 2x3 – 3 at the points where x = 2 and x = – 2 are parallel.

Asked by Topperlearning User | 7th Aug, 2014, 08:27: AM

Expert Answer:

The equation of the curve is y = 2x3 – 3     …(i)

Differentiating w.r.t. x, we get
dy over dx equals 6 straight x squared 
Now,   m1 = Slope of the tangent at  x equals 2 space comma space open parentheses dy over dx close parentheses subscript x equals 2 end subscript equals 6 cross times left parenthesis 2 right parenthesis squared equals 24
and,     m2 = Slope of the tangent at straight x equals minus 2 comma space open parentheses dy over dx close parentheses subscript straight x equals minus 2 end subscript equals 6 cross times left parenthesis minus 2 right parenthesis squared equals 24
Clearly, m1 = m2
Thus, the tangents to the given curve at the points where x = 2 and x = –2 are parallel.

Answered by  | 7th Aug, 2014, 10:27: AM