What is the equation of the tangent line of #f(x) =1-x^2/(x-5)# at # x = 4#?

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Answer 1 · 2017-02-10T03:49:28

#24x-y-73=0#. See the tangent-inclusive Socratic graph.

graph{(-x-6-25/(x-5)-y)(24x-y-81)=0 [0, 4.4, 5.808, 14.8, 15.2]}

By actual division,

y=-x-6-25/(x-5) = 15, at x = 4.

So, the point of contact of the tangent is P( 4, 15 ).

The slope m of the tangent is fi at x = 4.

f#'=-1+25/(x-5)^2=24 =m.

So, the equation to the tangent at P is

#y-15=24(x-4)#, giving

#24x-y-81=0#