Question

1. graph the piecewise-defined function. ( y = f(x) = \begin{cases} x + 6 & \text{if } x < 0 \\ 6 - & \text{if } x geq 0 end{cases} )

Explanation:

Step1: Analyze the first piece ($x < 0$)

The first part of the piecewise function is $y = x + 6$ for $x < 0$. This is a linear function with a slope of $1$ and a y - intercept of $6$. When $x = 0$, $y=0 + 6=6$, but since $x < 0$, the point $(0,6)$ is not included for this part. We can find another point, for example, when $x=- 6$, $y=-6 + 6 = 0$, so the point $(-6,0)$ is on this line segment. The graph of this part is a line with slope $1$ starting from the left (as $x$ approaches $-\infty$) and approaching the point $(0,6)$ (but not including it).

Step2: Analyze the second piece ($x\geq0$)

Assuming the second part is $y = 6 - x$ (since the original has a typo, and it should be a linear function for $x\geq0$ to make sense for graphing a piecewise function). For $y=6 - x$ with $x\geq0$, the slope is $- 1$ and the y - intercept is $6$. When $x = 0$, $y=6-0 = 6$, so the point $(0,6)$ is included here. When $x = 6$, $y=6 - 6=0$, so the point $(6,0)$ is on this line segment. The graph of this part is a line with slope $-1$ starting from the point $(0,6)$ and going to the right (as $x$ approaches $+\infty$).

Step3: Graphing the function

• For the interval $x < 0$: Plot the line $y=x + 6$. Use an open circle at $(0,6)$ (because $x = 0$ is not in the domain of this part) and draw the line going to the left with a slope of $1$.
• For the interval $x\geq0$: Plot the line $y = 6 - x$. Use a closed circle at $(0,6)$ (because $x = 0$ is in the domain of this part) and draw the line going to the right with a slope of $-1$.

(Note: Since the problem had a typo in the second part of the piecewise function, we assumed a reasonable linear function $y = 6 - x$ for $x\geq0$ to proceed with graphing. If the intended function for $x\geq0$ is different, the graphing steps would be adjusted accordingly.)

Answer:

The graph consists of two line segments:

• For $x < 0$, the line $y=x + 6$ (open circle at $(0,6)$) with slope $1$.
• For $x\geq0$, the line $y = 6 - x$ (closed circle at $(0,6)$) with slope $- 1$.