Problem 3, part 1
The graph is shown below. Refer to figure 1. It looks like you have the right idea, but the left-most piece of the graph is incorrect. Take note how the left-most piece is going uphill when moving left to right. This is due to the positive slope for y = 2x+5.
The graph shows that nearly the entire real number line is covered when it comes to the range; however, there's a gap from y = 4 to y = 6. We're not including y = 4 in the range, but we include y = 6 in the range.
The lower part of the range spans from negative infinity up to y = 4, excluding that endpoint. So we have [tex](-\infty, 4)[/tex] as part of the range.
The upper half of the range is from y = 6 onward to infinity. We include y = 6. This means [tex][6, \infty)[/tex] is also part of the range.
Answer: The range in interval notation is [tex](-\infty, 4) \ \cup \ [6, \infty)[/tex]
The square bracket says to include 6 as part of the range. The "U" symbol is the union operation to glue together the two disjoint intervals.
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Problem 3, part 2
Move onto figure 2 for the graph. It appears you didn't graph the V shape correctly for the absolute value piece of y = 2|x+6|. That's the red portion of the graph in figure 2. As the graph shows, we can target any nonnegative y value. Therefore y = 0 or larger. In terms of an inequality, we would say the range is [tex]y \ge 0[/tex]
Answer: The range in interval notation is [tex][0, \infty)[/tex]
