Exercise 47 Page 513

Consider the general equation of a reciprocal function. Let's start by graphing the parent function to all reciprocal functions, $y=\frac{1}{x}.$ This function has a constant of variation of $1,$ and a horizontal and vertical asymptote of $0.$

Notice that the minimum distance from the origin to this function is a $45^{\circ }$ line from $(0,0)$ to $(1,1)$ or $(\text{-}1,\text{-}1).$ Let's add this to the graph.

If we zoom in on one of the quadrants, we notice that the distance is the hypotenuse of a $45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }$ triangle. In such a triangle the hypotenuse is always $\sqrt{2}$ longer than any of the legs. Since the legs are both $1$ unit long, the hypotenuse must be $\sqrt{2}$ units.

As we can see, when the constant of variation is $1$ the function passes through $(1,1).$ If we can push this graph outward so that the closest point to the origin is $(4,4),$ we will have a reciprocal function with a minimum distance of $4\sqrt{2}$ units. This is because this point creates a $45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }$ triangle with a leg length of $4.$

To determine the function we should substitute the known point and solve for $k.$ One function is $y=\frac{16}{x}.$ By placing a negative sign before $\frac{16}{x}$ we get a second function with the same characteristics.