To graph the piecewise-defined function
\[ f(x) = \begin{cases} x + 2 & \text{if } x < -1 \ -2x + 3 & \text{if } x \geq -1 \end{cases} \]
we will handle each piece separately.
Step 1: Graph the first piece \( f(x) = x + 2 \) for \( x < -1 \)
- This is a straight line with a slope of 1 and a y-intercept of 2.
- We only need to consider the portion of this line to the left of \( x = -1 \) (not including \( x = -1 \)).
- To find the endpoint at \( x = -1 \), substitute \( x = -1 \): \[ f(-1) = -1 + 2 = 1 \]
- Since this piece does not include \( x = -1 \), we will plot this point with an open circle at (-1, 1).
Step 2: Graph the second piece \( f(x) = -2x + 3 \) for \( x \geq -1 \)
- This is also a straight line but has a slope of -2 and a y-intercept of 3.
- We need to start this line at \( x = -1 \) where we will have a closed circle (since \( x = -1 \) is included).
- Calculate the value at \( x = -1 \) to find the point: \[ f(-1) = -2(-1) + 3 = 2 + 3 = 5 \]
- Thus the point at \( (-1, 5) \) will be closed.
Now, we can sketch the graph:
- Draw the line segment for \( f(x) = x + 2 \) starting from the y-intercept (0, 2) and extending to the left until just before (-1, 1). Use an open circle at (-1, 1).
- Draw the line segment for \( f(x) = -2x + 3 \) starting at the point (-1, 5) and extending rightward. You can choose another point to help graph this line, for example:
- If \( x = 0 \): \[ f(0) = -2(0) + 3 = 3 \quad \text{(point (0, 3))} \]
- Connect these points with a line, ensuring to leave appropirate closure for (-1, 1) as an open circle and (-1, 5) as a closed circle.
This gives you the piecewise function graph with the specified characteristics.
Final Graph Details:
- The line \( x + 2 \) will extend indefinitely to the left, starting from the y-intercept (0, 2) down to just before (-1, 1).
- The line \( -2x + 3 \) begins at (-1, 5) and includes that point and continues downward to the right.
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