Answer :
To graph the function [tex]\( h \)[/tex] defined as:
[tex]\[ h(x) = \begin{cases} x & \text{if } x \neq 2 \\ 3 & \text{if } x = 2 \end{cases} \][/tex]
we will proceed through the following steps:
1. Understand the Function:
- For all values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex], the function [tex]\( h(x) \)[/tex] is linear and defined as [tex]\( h(x) = x \)[/tex].
- Exactly at [tex]\( x = 2 \)[/tex], the function [tex]\( h(x) \)[/tex] takes the value [tex]\( h(2) = 3 \)[/tex], which is a point discontinuity (it differs from the value [tex]\( h(x) = x \)[/tex] that would normally be [tex]\( h(2) = 2 \)[/tex]).
2. Graph the Linear Part:
- Plot the line [tex]\( y = x \)[/tex] for all [tex]\( x \neq 2 \)[/tex]. This line passes through the origin [tex]\((0, 0)\)[/tex] and has a slope of 1 (i.e., for each unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] also increases by 1).
3. Mark the Point of Discontinuity:
- At [tex]\( x = 2 \)[/tex], instead of following the line [tex]\( y = x \)[/tex], the function [tex]\( h(x) \)[/tex] jumps to [tex]\( h(2) = 3 \)[/tex].
- Represent this point [tex]\((2, 3)\)[/tex] with a filled circle to indicate that the function takes this value at [tex]\( x = 2 \)[/tex].
- Draw an open circle at the point [tex]\((2, 2)\)[/tex] on the line [tex]\( y = x \)[/tex] to illustrate that the value [tex]\( x = 2 \)[/tex] is not included in the linear part of the function.
4. Add Axes and Labels:
- Draw and label the [tex]\( x \)[/tex]-axis and [tex]\( y \)[/tex]-axis.
- Provide axis labels and appropriate tick marks for reference.
### Detailed Steps to Plot:
1. Draw the Line [tex]\( y = x \)[/tex]:
- For [tex]\( x \)[/tex] ranging from a wide interval like [tex]\([-10, 10]\)[/tex]:
- When [tex]\( x = -10 \)[/tex], [tex]\( y = -10 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 10 \)[/tex]
- Connect these points with a straight line, but exclude the point where [tex]\( x = 2 \)[/tex].
2. Add the Special Points:
- Draw an open circle at [tex]\((2, 2)\)[/tex] to indicate that [tex]\( h(x) \)[/tex] does not equal 2 when [tex]\( x = 2 \)[/tex].
- Put a filled circle at [tex]\((2, 3)\)[/tex] to show that [tex]\( h(2) = 3 \)[/tex].
### Final Graph:
1. Draw the [tex]\( y = x \)[/tex] Line Excluding [tex]\( x = 2 \)[/tex]:
- A diagonal line passing through the origin (0,0) and continuing in both positive and negative directions, with a gap at [tex]\( x = 2 \)[/tex].
2. Mark [tex]\( (2, 3) \)[/tex] with a Filled Circle:
- This distinct point signifies the value of the function at [tex]\( x = 2 \)[/tex].
3. Relevant Labels:
- Label the vertical axis (y-axis) and horizontal axis (x-axis).
- Optionally, label specific points or add grid lines for clarity.
This graph will illustrate both the continuous linear behavior of [tex]\( h \)[/tex] away from [tex]\( x = 2 \)[/tex] and the unique behavior of [tex]\( h \)[/tex] at [tex]\( x = 2 \)[/tex].
[tex]\[ h(x) = \begin{cases} x & \text{if } x \neq 2 \\ 3 & \text{if } x = 2 \end{cases} \][/tex]
we will proceed through the following steps:
1. Understand the Function:
- For all values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex], the function [tex]\( h(x) \)[/tex] is linear and defined as [tex]\( h(x) = x \)[/tex].
- Exactly at [tex]\( x = 2 \)[/tex], the function [tex]\( h(x) \)[/tex] takes the value [tex]\( h(2) = 3 \)[/tex], which is a point discontinuity (it differs from the value [tex]\( h(x) = x \)[/tex] that would normally be [tex]\( h(2) = 2 \)[/tex]).
2. Graph the Linear Part:
- Plot the line [tex]\( y = x \)[/tex] for all [tex]\( x \neq 2 \)[/tex]. This line passes through the origin [tex]\((0, 0)\)[/tex] and has a slope of 1 (i.e., for each unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] also increases by 1).
3. Mark the Point of Discontinuity:
- At [tex]\( x = 2 \)[/tex], instead of following the line [tex]\( y = x \)[/tex], the function [tex]\( h(x) \)[/tex] jumps to [tex]\( h(2) = 3 \)[/tex].
- Represent this point [tex]\((2, 3)\)[/tex] with a filled circle to indicate that the function takes this value at [tex]\( x = 2 \)[/tex].
- Draw an open circle at the point [tex]\((2, 2)\)[/tex] on the line [tex]\( y = x \)[/tex] to illustrate that the value [tex]\( x = 2 \)[/tex] is not included in the linear part of the function.
4. Add Axes and Labels:
- Draw and label the [tex]\( x \)[/tex]-axis and [tex]\( y \)[/tex]-axis.
- Provide axis labels and appropriate tick marks for reference.
### Detailed Steps to Plot:
1. Draw the Line [tex]\( y = x \)[/tex]:
- For [tex]\( x \)[/tex] ranging from a wide interval like [tex]\([-10, 10]\)[/tex]:
- When [tex]\( x = -10 \)[/tex], [tex]\( y = -10 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 10 \)[/tex]
- Connect these points with a straight line, but exclude the point where [tex]\( x = 2 \)[/tex].
2. Add the Special Points:
- Draw an open circle at [tex]\((2, 2)\)[/tex] to indicate that [tex]\( h(x) \)[/tex] does not equal 2 when [tex]\( x = 2 \)[/tex].
- Put a filled circle at [tex]\((2, 3)\)[/tex] to show that [tex]\( h(2) = 3 \)[/tex].
### Final Graph:
1. Draw the [tex]\( y = x \)[/tex] Line Excluding [tex]\( x = 2 \)[/tex]:
- A diagonal line passing through the origin (0,0) and continuing in both positive and negative directions, with a gap at [tex]\( x = 2 \)[/tex].
2. Mark [tex]\( (2, 3) \)[/tex] with a Filled Circle:
- This distinct point signifies the value of the function at [tex]\( x = 2 \)[/tex].
3. Relevant Labels:
- Label the vertical axis (y-axis) and horizontal axis (x-axis).
- Optionally, label specific points or add grid lines for clarity.
This graph will illustrate both the continuous linear behavior of [tex]\( h \)[/tex] away from [tex]\( x = 2 \)[/tex] and the unique behavior of [tex]\( h \)[/tex] at [tex]\( x = 2 \)[/tex].
