What is the equation of the tangent line of #f(x) = (4x^2-x+3)/(x-1)# at #x=2#?

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Answer 1 · 2017-08-30T11:59:34

Equation of tangent at # x=2 # is # 2x +y= 21#

#f(x) = (4x^2-x+3)/(x-1) ; x=2 # , When #x=2#

#f(x) = (4*2^2-2+3)/(2-1)= 17 # . So we want thagent at

point # (2,17)# . Slope of the curve at # (2,17)# is

#f^'(x) :. f^'(x) = (dy/dx(4x^2-x+3)* (x-1)-dy/dx(x-1) *(4x^2-x+3))/(x-1)^2#

(quotient rule). #f^'(x)= ((8x-1) * (x-1) - 1 * (4x^2-x+3))/(x-1)^2 #.

At #x=2, f^'(x) = ((8*2-1) *(2-1) - (4*2^2-2+3))/((2-1)^2# or

# f^'(x) = (15 * 1 - 17)/1= -2 :.# slope of tangent at point

# (2,17)# is #m=-2#. Equation of tangent at # (2,17)# is

#(y-y_1)=m(x-x_1) or y- 17 = -2 ( x-2) # or

# y-17 = -2x +4 or 2x +y= 21# [Ans]