Question

1. $f(x)=\

$$\begin{cases}-x + 3&\\text{if }x < 2\\\\2x - 3&\\text{if }x\\geq2\\end{cases}$$

$

Explanation:

Step1: Analyze the first piece of the function

For \( y=-x + 3\) (\(x<2\)):

• When \(x = 0\), \(y=-0 + 3=3\). So we have the point \((0,3)\).
• When \(x=2\) (approaching from the left), \(y=-2 + 3 = 1\).

Step2: Analyze the second piece of the function

For \(y = 2x-3\) (\(x\geq2\)):

• When \(x = 2\), \(y=2\times2-3=1\).
• When \(x=3\), \(y=2\times3 - 3=3\).

Step3: Plot the points and draw the lines

• For \(y=-x + 3\) (\(x<2\)): Plot \((0,3)\) and use the slope \(m=-1\) (since \(y=-x + 3\) is in \(y=mx + b\) form with \(m=-1\) and \(b = 3\)). Draw a line (with an open - circle at \(x = 2\) because \(x<2\)).
• For \(y=2x-3\) (\(x\geq2\)): Plot \((2,1)\) and \((3,3)\). Use the slope \(m = 2\) (since \(y=2x-3\) is in \(y=mx + b\) form with \(m = 2\) and \(b=-3\)). Draw a line (with a closed - circle at \(x = 2\) because \(x\geq2\)).

Answer:

The graph of the piece - wise function \(f(x)=

$$\begin{cases}-x + 3&x<2\\2x-3&x\geq2\end{cases}$$

\) is composed of two line segments. One is \(y=-x + 3\) for \(x<2\) (open - circle at \((2,1)\)) and the other is \(y=2x-3\) for \(x\geq2\) (closed - circle at \((2,1)\)).