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What is the equation of the tangent line of #f(x) =(2x-3)/(x-8)^2# at #x=3/2#?

Answer 1

Christopher Allers

#169y-8x+12=0

As #f(x)=(2x-3)/(x-8)^2# at #x=3/2#, #f(x)=0#

Hence, we are seeking tangent at #(3/2,0)#.

As slope of tangent is given by #f'(x)#, using quotient rule we have

#f'(x)=(2(x-8)^2-2(2x-3)(x-8))/(x-8)^4#

= #(-2x^2+6x+80)/(x-8)^4#

and when #x=3/2#, #f'(x)=(-2*9/4+6*3/2+80)/(3/2-8)^4#

= #169/2xx(2/13)^4=8/169#

and hence equation of tangent is

#y=8/169(x-3/2)# or #169y-8x+12=0#

Graph shown below - not drawn to scale

graph{(169y-8x+12)(y-(2x-3)/(x-8)^2)=0 [-0.5, 3., -0.3, 0.3]}

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Answer 2

Christopher Allers

The equation of the tangent line of f(x) = (2x-3)/(x-8)^2 at x=3/2 is y = -1/2x + 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.