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The tangent of f(x) = (x-9)/(5x + 4) would be the asymptote which is x = -1... 

f(x) = (x-9)/(5x+4)    Find an equation for the tangent line of f at x=8

$f(8)= \frac{-1}{44}$

$f(x) = \frac{(x-9)}{(5x+4)}\\ f'(x)=\frac{(5x+4)(1)-5(x-9)}{(5x+4)^2} \\ f'(x)=\frac{5x+4-5x+45}{(5x+4)^2} \\ f'(x)=\frac{49}{(5x+4)^2} \\ f'(8)=\frac{49}{(44)^2} \\ f'(x)=\frac{49}{1936}$

Equ of tangent

$\frac{49}{1936}=\frac{y+\frac{1}{44}}{x-8}\\ 49(x-8)=1936(y+\frac{1}{44})\\ 49x-392=1936y+44\\ 49x-1936y=436$

I suggest you check this carefully