Answer
$\text{(a)}$ $(-2,\infty)$.
$\text{(b)}$ $x$-intercetp $=0$; $y$-intercept $=0$
$\text{(c)}$ See graph below.
$\text{(d)}$ $(-6,\infty)$.
$\text{(e)}$ not continuous
Work Step by Step
Use the given piece-wise function, we can find:
$\text{(a)}$ The $x$-values of the function are the real numbers greater than $-2$. Hence, the domain $(-2,\infty)$.
$\text{(b)}$ For $x$-intercept, let $f(x)=0$ then solve for $x$.
$0=3x \longrightarrow x=0$, while $0=x+1\longrightarrow x=-1$.
Note that $f(0)=0$ and $f(-1) = -3$.
Thus, the $x$-intercept is $0$.
For $y$-intercept, let $x=0$ then solve for $y$.
Note that $f(0)=3(0)=0$,
Thus, the $y$-intercept is $0$.
$\text{(c)}$ Graph each part/piece of the function..
For $f(x)=3x, -2\le x \le 1$:
When $x=-2$, $f(-2)= 3(-2)=-6$. However, since $-2$ is not part of the domain, then plot the point $(-2, -6)$using a hollow dot. When $x=1$, $f(1)=3(1)=3$. Thus, the graph contains $(1, 3)$.
Plot the points $(-3, -6)$ and $(1, 3)$ then connect the points using a solid line.
For $f(x)=x+1, x> 1$:
When $x=1$, $f(1)=1+1=2$. However, since $1$ is not part of the domain, then plot the point $(1, 2)$using a hollow dot. When $x=5$, $f(5)=5+1=6$. Thus, the graph contains $(5, 6)$.
Plot the points $(1, 2)$ and $(5, 6)$ then connect them using a line, and finally, extend the line by putting an arrowhead going upward to the right.
Refer to the graph below.
$\text{(d)}$ The graph shows that the value of $y$ is any real number greater then $-6$. Hence, the range is $(-6,\infty)$.
$\text{(e)}$ The function is not continuous on its domain (discontinuous at $x=1$).
