Question

1. $f(x)=\

$$\begin{cases}-x + 2&\\text{if }x < 0\\\\\\sqrt{x}+3&\\text{if }x\\geq0\\end{cases}$$

$

Explanation:

Step1: Analyze the piece - wise function for \(x < 0\)

For \(x<0\), the function is \(y=-x + 2\). When \(x = 0\), \(y=2\) (but this point is not included in this part of the piece - wise function). We can find another point, for example, when \(x=-2\), \(y=-(-2)+2=4\).

Step2: Analyze the piece - wise function for \(x\geq0\)

For \(x\geq0\), the function is \(y = \sqrt{x}+3\). When \(x = 0\), \(y=\sqrt{0}+3=3\). When \(x = 1\), \(y=\sqrt{1}+3=4\), when \(x = 4\), \(y=\sqrt{4}+3=5\)

To graph the function:

• For \(y=-x + 2\) with \(x<0\), we draw a ray starting from the open - circle at \((0,2)\) and going to the left using the slope \(m=-1\).
• For \(y=\sqrt{x}+3\) with \(x\geq0\), we note that the domain of \(y = \sqrt{x}\) is \(x\geq0\) and we shift the graph of \(y=\sqrt{x}\) up by \(3\) units. We start with the point \((0,3)\) (closed - circle) and then plot other points like \((1,4)\) and \((4,5)\) and draw the curve for \(x\geq0\).

Answer:

Graph the ray \(y=-x + 2\) with \(x<0\) (open - circle at \((0,2)\)) and the curve \(y=\sqrt{x}+3\) with \(x\geq0\) (closed - circle at \((0,3)\)).