Answer
The equation of the tangent line is $y=-\frac{1}{2}x+2$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=8; u'(x)=0$
$v(x)=x^2+4; v'(x)=2x$
$f'(x)=\frac{(0)(x^2+4)-(2x)(8)}{(x^2+4)^2}=-\frac{16x}{(x^2+4)^2}$.
$f'(2)=-\frac{32}{64}=-\frac{1}{2}$.
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-1)=-\frac{1}{2}(x-2)\rightarrow y=-\frac{1}{2}x+2$.
