## Precalculus (6th Edition) Blitzer

(a) The domain of the function is defined as the set of possible values of x that can be put in the function to get a defined value. It can be easily determined by observing the plot of the function and observing the values of x for which the function graph exists. From the graph, one can see that the function is expanding in both directions of $x$ as there are arrows in both ends of the graph. Thus, the domain of the graph is $\left( -\infty ,\infty \right)$. (b) The range of the function is defined as the set of possible values of y that can be obtained in the function by a definite value of x. It can be easily determined by observing the plot of function and observing the values of y the function obtains in the domain. From the graph, we see that the function is expanding in the positive direction of $y$ and is limited to $-4$ in the negative direction of $y$. Hence, the range of the graph is $[-4,\infty )$. (c) The x intercept of the function is defined as the x coordinate of the point where the function meets the x axis. It can be determined by simply observing the plot and observing the coordinates of the point of intersection of the graph with the x axis. From the graph, we see that the function is intersecting the $x$-axis at $-3$ and $1$. Hence, the $x$ -intercepts of the function are $-3$ and $1$. (d) The y intercept of the function is defined as the y coordinate of the point where the function meets the y axis. It can be determined by simply observing the plot and observing the coordinates of the point of intersection of the graph with the y axis. From the graph, we see that the function is intersecting the $y$ -axis at $-3$. Hence, the $y$ -intercept of the function is $-3$. (e) Let us mark the point 2 on the x axis in the graph of the function. The value of y denotes the value of the function at $x=2$. Observe the y coordinate at $x=2$ to get the value of the function. We see that the value is $5$. Therefore, $f\left( 2 \right)=5$.