What is the equation of the line tangent to #f(x)=(x-1)^2-3x^2 # at #x=4#?

1 Answer
Apr 27, 2018

#y = -16x + 25#

Explanation:

The tangent is the line that just includes one point of the curve. That line is the first derivative of the curve at that point.
#f(x)=(x−1)^2−3x^2 = x^2 - 2x +1 - 3x^2#
#f(x) = -2x^2 - 2x +1#
#f'(x) = -4x - 2#

at x = 4, #f'(x) = -16# This is the slope of the desired line.

The equation of a line can be written in the form
#(y−y_0)=m(x−x_0)#, where m represents the slope of the line, and #(x_0,y_0)# represents a point on the line.

Our original curve yields the point (4, -39)
#f(4) = -2x^2 - 2x +1 = -39#

The line equation is thus:
#(y− (-39)) = -16(x−4)# ; #y + 39 = -16x + 64#
#y = -16x + 25#