How do you find the equation of a line tangent to the function #y=sqrtx+1# at x=4?

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Answer 1 · 2016-07-24T05:52:00

#x-4y+8=0#

As #y=sqrtx+1# and we have to find tangent at #x=4#, complete coordinates of point where tangent is needed are #(4,sqrt4+1)# or #(4,3)#.

Now as slope of tangent is given by first derivative of function, we have slope as

#(dy)/(dx)=1/2xx x^(1/2-1)=1/(2sqrtx)#

and hence slope at #x=4# is #1/(2sqrt4)=1/4#

Hence given point #(4,3)# and slope #1/4#, equation of tangent is given by point slope form and is

#(y-3)=1/4(x-4)# or

#4y-12=x-4# or #x-4y+8=0#

graph{(y-sqrtx-1)(x-4y+8)=0 [-1.19, 8.81, 0.52, 5.52]}