How do you find the equation of a line tangent to the function #y=2-sqrtx# at (4,0)?

1 Answer
Oct 18, 2016

#y=(-1/4)x+1#

Explanation:

The #color(red)(slope)# of the tangent line to the given function #2-sqrtx# is #color(red)(f'(4))#

Let us compute #color(red)(f'(4))#
#f(x)=2-sqrtx#
#f'(x)=0-1/(2sqrtx)=-1/(2sqrtx)#

#color(red)(f'(4))=-1/(2sqrt4)=-1/(2*2)=color(red)(-1/4)#

Since this line is tangent to the curve at #(color(blue)(4,0))#
then it passes through this point :
Equation of the line is :

#y-color(blue)0=color(red)(-1/4)(x-color(blue)4)#
#y=(-1/4)x+1#