How do find the equation of the tangent of the curve of #y = x^2 - 4x# at the point with coordinates (2,-4)?

1 Answer
Apr 8, 2015

You can use the derivative to determine the slope of the tangent and then use the slope-point form for the equation.

#y = x^2 - 4x#

#(dy)/(dx) = 2x - 4#

when #x=2# this gives a slope of 2(2) - 4 = 0
(Actually at this point we know the tangent is a horizontal line,
#y = -4# but we'll continue on to demonstrate the method).

Using the slope #m = 0# and the point #(2,-4)#
the slope point form, #(y-y_1) = m(x-x_1)#
becomes
#y-(-4) = (0) (x -2)#

#rarr y = -4#