Answer
a) $m= -\dfrac{1}{8}$
b) $ y=-\dfrac{1}{8}x-1$
Work Step by Step
We have: $(xy)^2+xy-x=8$
We differentiate both sides with respect to $x$.
$2x^2 y\dfrac{dy}{dx}+x \dfrac{dy}{dx}=1 -y-2xy^2\\ \dfrac{dy}{dx}=\dfrac{1-y-2xy^2}{2x^2y+x} $
a) The slope at $( -8, 0)$ is: $m=\dfrac{1-0-0}{0-8} = -\dfrac{1}{8}$
b) The equation of a tangent line at $( -8,0)$ is:
$y-y_1=m(x-x_1)\\ y -0=-\dfrac{1}{8} (x+8)\\ y=-\dfrac{1}{8}x-1$
