Question

Graph the following function, using open circles for points on the ends of lines that are not in the solution set, and solid points for those that are. $$ \begin{array}{ll} y=x+2 & x \leq-4 \\ y=\frac{1}{4} x^2 & -45 \end{array} $$

Short answer

Graph involves plotting points and curves for four different function segments, combining them as described, with correct open/solid circles at endpoints.

Step-by-step solution

Analyze the first function segment

The first part of the function is given as \( y = x + 2 \) for \( x \leq -4 \). Identify the endpoint: when \( x = -4 \), \( y = -4 + 2 = -2 \). Since \( x \leq -4 \), draw a solid point at (-4, -2). Graph the line with open endpoints extending left from this point.

Analyze the second function segment

The second part of the function is \( y = \frac{1}{4} x^2 \) for \( -4 < x < 0 \). At \( x = -4 \), the value is \( y = \frac{1}{4}(-4)^2 = 4 \); use an open circle at (-4, 4). At \( x = 0 \), \( y = \frac{1}{4}(0)^2 = 0 \); use another open circle at (0, 0). Plot additional points (e.g. when \( x = -2 \), \( y = 1 \)) and draw the parabolic curve between these points.

Analyze the third function segment

The third function is \( y = \frac{6}{25} x^2 \) for \( 0 \leq x \leq 5 \). At \( x = 0 \), \( y = 0 \); use a solid point at (0, 0). At \( x = 5 \), \( y = \frac{6}{25}(5)^2 = 6 \); use a solid point at (5, 6). Plot additional points (e.g. when \( x = 2 \), \( y = \frac{6}{25}(2)^2 = 0.96 \)) and draw the parabolic curve.

Analyze the fourth function segment

The final part of the function is \( y = -2x + 6 \) for \( x > 5 \). Identify the endpoint at \( x = 5 \), \( y = -2(5) + 6 = -4 + 6 = 6 \); since \( x > 5 \), draw an open circle at (5, 6). Graph the line extending right from this open endpoint.

Combine the segments

Combine all the points and curves from each segment. Ensure you use solid circles where indicated and open circles where necessary. Confirm the graph matches the given domain conditions for each segment.

Key concepts

Graphing Functions

Graphing piecewise functions is like piecing together parts of a puzzle. Each part of the function applies to a specific range of the input values (x-values) and can be a different type of graph, like a line or a parabolic curve.
This exercise has four different segments:

• Linear equation for \(x \leq -4\)
• Parabolic curve for \(-4 < x < 0 \)
• Another parabolic but different equation for \(0 \leq x \leq 5\)
• Another linear equation for \(x > 5\)

To graph these segments, follow these steps:

• Start with the first equation and plot all the points where the range applies
• Draw a solid point if the endpoint is included ( \[ \leq \text{ or } \geq \] ) and an open point if it is not
• Move to the next function and repeat until you have covered all segments

The key is to pay attention to the endpoints and connect the points smoothly where necessary.

Function Endpoints

In piecewise functions, endpoints are crucial because they tell us where one segment of the function ends and another begins. Here, we have to determine whether these endpoints are included in the graph.
Solid Points: Indicate that the endpoint is included in the function's solution.
Open Circles: Indicate that the endpoint is not included in the function's solution.
Let's see how this works in our segments:

• For \(y = x + 2\) when \(x = -4\), this endpoint is included, so we use a solid point.
• For \(y = \frac{1}{4} x^2\) when \-4 < x < 0, neither \-4 nor 0 are included, so we use open circles at these points.
• For \(y = \frac{6}{25} x^2\) when \(0 \leq x \leq 5\), both 0 and 5 are included, so we use solid points.
• For \(y = -2x + 6\) when \(x > 5\), the point at \(x = 5\) is not included, so we use an open circle.

Putting the correct type of endpoint helps give an accurate graph.

Parabolic Curves

The given exercise includes two parabolic curve sections. Parabolas are U-shaped graphs that can open upwards or downwards based on the equation. For the parabolic segments in the exercise:

• The first parabolic segment is \(\frac{1}{4} x^2\), which opens upwards because the coefficient \( \frac{1}{4} \) is positive.
• The second parabolic segment is \(\frac{6}{25} x^2\), which also opens upwards as \( \frac{6}{25} \) is positive.

The key steps to graph these are:

• Select a few key points within the range.
• Plot these points to see the general U-shape.
• Ensure the ends of the range are correctly marked with either solid or open circles.

In both cases, remember to plot additional points to ensure the curve is accurate. For instance, in the first segment \( -4 < x < 0 \), besides endpoints, you plot points like \(x = -2\). Doing so ensures your parabolic curves are smooth and precise between the specified intervals.