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 | |
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\(x = 1\) 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 \(x \neq 0\) | \(y = (x-1)^2\) | \(y = 2 + (x-1)^4\) | \(y = x\) |