The line #y=2x + k#, where k is a constant, is a tangent to the curve #y= -1+2x -x^2#. Find (a) the value of k; (b) the coordinate of their point of contact.?

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Answer 1 · 2018-02-08T06:29:03

#k = -1#

The point of contact is: #(0,-1)#

Compute the first derivative of the curve:

#dy/dx = 2-2x#

The slope of the line is 2, therefore, we set the first derivative equal to 2 and then solve for x:

#2 = 2-2x#

#x = 0 larr# this is the x coordinate of the point of contact.

The y coordinate of the point of contact is found by evaluating the function at x = 0:

#y = -1+2(0)-0^2#

#y = -1 larr# this is the y coordinate of the point of contact.

The point of contact is: #(0,-1)#

Find the value of k by evaluating the line at the point of contact:

#-1 = 2(0)+k#

#k = -1#