How do you find the equation of the line tangent to #f(x)= (sqrtx+1)#, at (0,1)?

1 Answer
Jan 26, 2016

Find the derivative at the point (= the slope of the graph's tangent). Then find the equation with the slope and point that you have.

Explanation:

  • First: the slope=the derivative= #d/dx[f(x)]#
    #d/dx[sqrt(x)+1]=d/dx[x^(1/2)+1]#

#=1/2*x^(-1/2)# #=1/2*1/sqrt(x)#

#=1/(2sqrt(x))#

  • The slope of the line tangent to f(x) at #(0,1)=#
    #1/(2sqrt(0))=1/0=>#the line is parallel to the #y#-axis

  • Second: the equation of the tangent is
    #x=a# where #a# is constant
    again it passes through (0,1)
    so equation should be #x=0# i.e. #y# axis