# Question bc921

The denominator factorises into $\left(x - 3\right) \left(x + 3\right)$ so there vertical asymptotes at $x = + 3$ and $x = - 3$. For large values of $x$ both positive and negative the expression is nearly $1$. (Alternatively, the expression=(x^2-9+9)/(x^2-9)=1+9/(x^2-9, which is clearly positive whenever $| x | > 3$ and tends to $1$ as $x$ tends to ± ∞#.) Replacing $x$ with $- x$ leaves the expression unchanged, so the graph is symmetrical about the $y$-axis.