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#