How do you find an equation of the tangent line to the curve at the given point #y=sinx+3# and #x=pi#?

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Answer 1 · 2017-01-25T08:36:42

#y+x=3+pi#

we use

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

where #(x_1,y_1) # is a known coordinate; and #m=#gradient

#y=sinx+3#

#y(pi)=sinpi+3=0+3=3#

#(x_1,y_1) =(pi, 3)#

gradient of tangent #m=((dy)/(dx))_(x=pi)#

#y=sinx+3#

#(dy)/(dx)=cosx#

#m=((dy)/(dx))_(x=pi)=cospi=-1#

eqn tgt

#y-3=-1(x-pi)#

#y-3=-x+pi#

#y+x=3+pi#