Determine point A:
[tex]y(x)=\frac{10}{2x+1}-2\\\frac{10}{2x+1}-2=0\implies x=2[/tex]
so A(2,0)
Find the function for the tangent:
[tex]\frac{dy}{dx}=-\frac{-20}{(2x+1)^2}[/tex]
the tangent function is a line
[tex]y=ax + b = \frac{dy}{dx}|_A\cdot x + b\\[/tex]
(with slope being the derivative evaluated at point A(2,0))
[tex]\frac{dy}{dx}|_A=-\frac{4}{5}[/tex] so a=-4/5.
Determine b by using what we know about A:
[tex]y=-\frac{4}{5}x+b\\0=-\frac{4}{5}\cdot 2+b\implies b=\frac{8}{5}[/tex]
so the function is
[tex]y = -\frac{4}{5}x+\frac{8}{5}\\5y + 4x = 8[/tex]
(i) this is demonstrated above
(ii) distance AC: we know A(2,0) and C(0,5/8), the distance is:
[tex]\sqrt{(0-2)^2+(\frac{5}{8}-0)^2}=\sqrt{281}/8\approx 2.1[/tex]
