How do you find the equation of the tangent line to the curve #y=(2x+1)/(x+2) # at the given point (1,1)?

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Answer 1 · 2015-05-01T02:14:07

We have, #y = (2x+1)/(x+2)#

Now, the derivative (using quotient rule)

#dy/dx = (2*(x+2) - (2x+1))/(x+2)^2 = 3/(x+2)^2#

Now #dy/dx# at #(1,1)# is the slope of the tangent to the curve at #(1,1)#

#m = (dy/dx)_(1,1) = 3/9 = 1/3#

Now, point-slope form a straight line is given by

#(y-y_1) = m*(x-x_1)#
Now, #(x_1,y_1) = (1,1)# and #m = 1/3#, substituting these values in above equation we get,

#y-1 = 1/3 * (x-1)# => #y = x/3 + 2/3# (Answer)