Answer
(a) $y=\frac{55x}{x-55}$
(b) decrease.
(c) increase sharply.
Work Step by Step
(a) Express y as a function of x, and graph the function.
$\frac{1}{y}=\frac{1}{F}-\frac{1}{x}=\frac{x-F}{Fx}=\frac{x-55}{55x}$ so $y=\frac{55x}{x-55}$
(b) What happens to the focusing distance y as the object moves far away from the lens?
We assume $x\gt F$, as $x$ increases, $y$ will decrease.
(c) What happens to the focusing distance y as the object moves close to the lens?
Again we assume $x\gt F$, when $x$ gets close to $F$, the function will get close to
a vertical asymptote, where the $y$ value will increase sharply.
