Many ways problem

## Solution

Here is one possible way to complete the table.

$y$-axis is an asymptote Is symmetrical about $x = 1$ Intersects $y$-axis at $3$ Passes through origin
$x = 1$ is a root $y = \dfrac{x-1}{x}$ for $x \neq 0$ $y = \vert x - 1 \vert$ $y = -3x + 3$ $y = x(1-x)$
Has exactly two roots. $y = -1 + \dfrac{1}{x^2}$ for $x \neq 0$ $y = x(x-2)$ $y = (x-1)^2 (x+3)$ $y = x(x+7)$
$x$-axis is an asymptote $y = \dfrac{1}{x}$ for $x \neq 0$ $y = \dfrac{1}{(x-1)^2}$ for $x \neq 1$ $y = \dfrac{3}{x+1}$ for $x \neq -1$ $y = \dfrac{x}{(x-1)^2}$ for $x \neq 1$
$y \to \infty$ as $x \to \infty$ $y = x + \dfrac{1}{x}$ for $x \neq 0$ $y = (x-1)^2$ $y = 2 + (x-1)^4$ $y = x$