What is the equation of the line tangent to #f(x)= -(3-2x)^2 # at #x=-2#?
1 Answer
Feb 2, 2016
y - 28-x - 7 = 0
Explanation:
To obtain the equation of the tangent , y-b = m(x-a) , we
require to find the gradient , m , and the point ( a,b).
f'(x) is the gradient of the tangent and f'(-2) will give it's
value. The x-coord is given, x = -2 and to find b , the y-coord
evaluate f(-2).
f(x) =
# - (3-2x)^2# differentiate using the
# color(blue)(" chain rule")#
#f'(x) =-2(3-2x) d/dx (3-2x) = -2(3-2x).(-2) = 4(3-2x)# and f'(-2) = 4(3+4) = 28 = m
b = f(-2)
# = - (3+4)^2 = - 49 # equation: y +49 = 28(x+2)
# rArr y-28x -7 = 0#
