What is the equation of the tangent line of #f(x) =sqrtx-15# at #x=225#?

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Answer 1 · 2016-05-30T14:51:07

#y=x/30-15/2#

We have an equation #y=sqrtx-15# and at the point #x=225#

compute for the slope m

#f' (x)=1/(2sqrtx)#

#m=f' (225)=1/(2sqrt225)=1/30#

Solve for the required point #(x_1, y_1)#

#y_1=sqrt(225)-15=0#

#(x_1, y_1)=(225, 0)#

Solve for the equation of the tangent line

#y-y_1=m(x-x_1)#

#y-0=1/30(x-225)#

#y=x/30-15/2#

God bless....I hope the explanation is useful.