Consider the function $f ( x) =\dfrac{x+1}{x+2}$.

Solution (a) The domain of $f$ consists of all real numbers except $-2;$ that is, the set $\{ x|x\neq -2\}$.

(b) $$\begin{eqnarray*} &&{\rm When}\ x=1, {\rm then}\ f ( 1 )\underset{\underset{\color{#0066A7}{x=1}}{\color{#0066A7}{\uparrow}}} {=} \dfrac{1+1}{1+2}=\dfrac{2}{3}. \hbox{The point } \left( 1, \dfrac{2}{3}\!\right) \hbox{ is on the graph of } f; \hbox{ the}\\ \end{eqnarray*}$$ point $\left( 1,\dfrac{1}{2}\!\right)$ is not on the graph of $f$.

(c) $$\begin{eqnarray*} &&\hbox{If } x=2, \hbox{then } f ( 2) \underset{\underset{\color{#0066A7}{x=2}}{\color{#0066A7}{\uparrow}}} {=} \dfrac{2+1}{2+2}=\dfrac{3}{4}. \hbox{The point } \left( 2, \dfrac{3}{4}\!\right) \hbox{ is on the graph of } f.\\ \end{eqnarray*}$$

(d) If $f( x) =2$, then $\dfrac{x+1}{x+2}=2.$ Solving for $x$, we find $$\begin{eqnarray*} x+1 &=&2( x+2) =2x+4 \nonumber \\ x &=&-3 \end{eqnarray*}$$

The point $( -3,2)$ is on the graph of $f$.

(e) The $x$-intercepts of the graph of $f$ occur when $y=0$. That is, they are the solutions of the equation $f( x) =0$. The $x$ -intercepts are also called the real zeros or roots of the function $f$.

9

The real zeros of the function $f( x) =\dfrac{x+1}{x+2}$ satisfy the equation $x+1=0$ or $x=-1$. The only $x$-intercept is $-1$, so the point $( -1,0)$ is on the graph of $f.$