Question

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

Answer 1

#169y-8x+12=0

Explanation:

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]}