Answer
$y=-\frac{3}{16}x-\frac{1}{2}$, as shown in the figure.
Work Step by Step
Given the curve as $y=1/x^3$ and a point $(-2,-1/8)$ on the curve, we can find the slope of the tangent line at this point as:
$m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to0}\frac{1/(-2+h)^3-1/(-2)^3}{h}=\lim_{h\to0}\frac{2^3-(2-h)^3}{-2^3h(2-h)^3}=\lim_{h\to0}\frac{12h-6h^2+h^3}{-2^3h(2-h)^3}=\lim_{h\to0}\frac{12-6h+h^2}{-8(2-h)^3}=\frac{12-0+0^2}{-8(2-0)^3}=-\frac{3}{16}$
The equation for the tangent line at $(-2,-1/8)$ can then be written as $y+1/8=-\frac{3}{16}(x+2)$ or $y=-\frac{3}{16}x-\frac{1}{2}$, as shown in the figure.
