Answer
The equation of the tangent is $ y=\frac{x}{2}+3$.
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=27; u'(x)=0$
$v(x)=x^2+9; v'(x)=2x$
$f'(x)=\frac{(0)(x^2+9)-(2x)(27)}{(x^2+9)^2}$
$=-\frac{54x}{(x^2+9)^2}$
$f'(-3)=-\frac{54(-3)}{((-3)^2+9)^2}=\frac{1}{2}$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-\frac{3}{2})=\frac{1}{2}(x+3)\rightarrow y=\frac{x}{2}+3 $.
