Answer
The equation of the tangent is $y'=-6x+31$.
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x+3; u'(x)=1$
$v(x)=x-3; v'(x)=1$
$f'(x)=\frac{(x-3)-(x+3)}{(x-3)^2}=-\frac{6}{(x-3)^2}$
$f'(4)=-\frac{6}{(4-3)^2}=-6$.
Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-7)=-6(x-4)\rightarrow y=-6x+31$.
A graphing calculator and a computer algebra system have been used to confirm these results.
