Draw the graph of
(i) `y = |x+2| + |x-3|` . (ii) `y = x+(x)/(|x|)`

Answer

To draw the graphs of the given functions, we will analyze each function step by step.

### (i) Graph of \( y = |x+2| + |x-3| \)

1. **Identify critical points**:
- The expression inside the absolute values changes at \( x = -2 \) and \( x = 3 \). These points will help us define intervals to analyze the function.

2. **Define intervals**:
- We will consider three intervals based on the critical points:
- \( x < -2 \)
- \( -2 \leq x < 3 \)
- \( x \geq 3 \)

3. **Evaluate the function in each interval**:
- **For \( x < -2 \)**:
- Here, \( |x+2| = -(x+2) \) and \( |x-3| = -(x-3) \).
- Thus, \( y = -(x+2) - (x-3) = -x - 2 - x + 3 = -2x + 1 \).

- **For \( -2 \leq x < 3 \)**:
- Here, \( |x+2| = x+2 \) and \( |x-3| = -(x-3) \).
- Thus, \( y = (x+2) - (x-3) = x + 2 + 3 = 5 \).

- **For \( x \geq 3 \)**:
- Here, \( |x+2| = x+2 \) and \( |x-3| = x-3 \).
- Thus, \( y = (x+2) + (x-3) = 2x - 1 \).

4. **Summarize the piecewise function**:
- The function can be summarized as:
\[ y = \begin{cases} -2x + 1 & \text{if } x < -2 \\ 5 & \text{if } -2 \leq x < 3 \\ 2x - 1 & \text{if } x \geq 3 \end{cases} \]

5. **Plot the graph**:
- For \( x < -2 \): The line \( y = -2x + 1 \) has a slope of -2 and intercepts the y-axis at \( (0, 1) \).
- For \( -2 \leq x < 3 \): The line is horizontal at \( y = 5 \).
- For \( x \geq 3 \): The line \( y = 2x - 1 \) has a slope of 2 and intercepts the y-axis at \( (0, -1) \).

### (ii) Graph of \( y = x + \frac{x}{|x|} \)

1. **Identify critical points**:
- The expression \( |x| \) changes at \( x = 0 \). This will help us define intervals.

2. **Define intervals**:
- We will consider two intervals:
- \( x < 0 \)
- \( x > 0 \)

3. **Evaluate the function in each interval**:
- **For \( x < 0 \)**:
- Here, \( |x| = -x \).
- Thus, \( y = x + \frac{x}{-x} = x - 1 \).

- **For \( x > 0 \)**:
- Here, \( |x| = x \).
- Thus, \( y = x + \frac{x}{x} = x + 1 \).

4. **Summarize the piecewise function**:
- The function can be summarized as:
\[ y = \begin{cases} x - 1 & \text{if } x < 0 \\ x + 1 & \text{if } x > 0 \end{cases} \]
- Note that \( y \) is not defined at \( x = 0 \).

5. **Plot the graph**:
- For \( x < 0 \): The line \( y = x - 1 \) has a slope of 1 and intercepts the y-axis at \( (0, -1) \).
- For \( x > 0 \): The line \( y = x + 1 \) has a slope of 1 and intercepts the y-axis at \( (0, 1) \).