To graph the function [tex]\( h(x) = |x+2| - 1 \)[/tex] using transformations of the absolute value function [tex]\( f(x) = |x| \)[/tex], we can follow these steps:
1. Understand [tex]\( f(x) = |x| \)[/tex]:
- This is the parent function of the absolute value, which is a V-shaped graph with the vertex at the origin [tex]\((0, 0)\)[/tex].
- The function [tex]\( f(x) = |x| \)[/tex] is defined as:
[tex]\[
f(x) = \begin{cases}
x & \text{if } x \geq 0 \\
-x & \text{if } x < 0
\end{cases}
\][/tex]
- It has a slope of 1 for [tex]\( x \geq 0 \)[/tex] and a slope of -1 for [tex]\( x < 0 \)[/tex].
2. Horizontal Shift:
- The function [tex]\( |x+2| \)[/tex] represents a horizontal shift of the graph of [tex]\( |x| \)[/tex].
- Inside the absolute value, [tex]\( x \)[/tex] is replaced by [tex]\( x+2 \)[/tex], which means we shift the graph to the left by 2 units.
- The vertex of the graph shifts from [tex]\((0,0)\)[/tex] to [tex]\((-2,0)\)[/tex].
3. Vertical Shift:
- The function [tex]\( |x+2| - 1 \)[/tex] represents a vertical shift.
- Subtracting 1 from the function [tex]\( |x+2| \)[/tex] shifts the graph downward by 1 unit.
- The vertex of the graph now moves from [tex]\((-2,0)\)[/tex] to [tex]\((-2, -1)\)[/tex].
4. Graphing [tex]\( h(x) = |x+2| - 1 \)[/tex]:
- Begin by plotting the vertex at [tex]\((-2, -1)\)[/tex].
- From the vertex, the graph will have the same V-shape as [tex]\( f(x) = |x| \)[/tex].
- For [tex]\( x \geq -2 \)[/tex]: the slope is 1, so the graph moves upward to the right.
- For [tex]\( x < -2 \)[/tex]: the slope is -1, so the graph moves downward to the left.
By following these steps, we can transform [tex]\( f(x) = |x| \)[/tex] to graph [tex]\( h(x) = |x+2| - 1 \)[/tex].
### Coordinate Plotting:
- For a precise numerical approach, consider evaluating specific points:
[tex]\[
\begin{align*}
\text{For } x = -4: & \quad h(-4) = |-4+2| - 1 = | -2 | - 1 = 2 - 1 = 1 \quad ( -4, 1 ) \\
\text{For } x = -2: & \quad h(-2) = |-2+2| - 1 = | 0 | - 1 = 0 - 1 = -1 \quad ( -2, -1 ) \quad \text{(vertex)} \\
\text{For } x = 0: & \quad h(0) = |0+2| - 1 = | 2 | - 1 = 2 - 1 = 1 \quad ( 0, 1 ) \\
\text{For } x = 2: & \quad h(2) = |2+2| - 1 = | 4 | - 1 = 4 - 1 = 3 \quad ( 2, 3 )
\end{align*}
\][/tex]
#### Points to remember:
- [tex]\((-4, 1)\)[/tex]
- [tex]\((-2, -1)\)[/tex]
- [tex]\( (0, 1)\)[/tex]
- [tex]\( (2, 3)\)[/tex]
### Final graph of [tex]\(h(x)\)[/tex]:
- Plot the evaluated points on the Cartesian plane.
- Connect these points to form the V-shaped graph.
- Ensure the slopes are consistent with the absolute value function's behavior (upward and downward straight lines from the vertex).
By carefully following these transformations and plotting, you can accurately graph [tex]\( h(x) = |x+2| - 1 \)[/tex] based on the described steps.
