Question

(b) the solution set in interval notation for |y + 1| > 0 is (-∞, -1) ∪ (-1, ∞).

part: 2 / 4

part 3 of 4

graph the solution set for |y + 1| > 0.

← -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 →

Explanation:

Step1: Analyze the inequality

The inequality is \(|y + 1|>0\). The absolute value of a number is greater than 0 when the number inside the absolute value is not equal to 0. So we solve \(y+1
eq0\), which gives \(y
eq - 1\).

Step2: Graph the solution

On the number line, we need to represent all real numbers except \(y=-1\).

• For the interval \((-\infty,-1)\), we draw an arrow starting from the left end of the number line (towards \(-\infty\)) up to but not including \(-1\). We use an open circle at \(-1\) to indicate that \(-1\) is not included.
• For the interval \((-1,\infty)\), we draw an arrow starting from just after \(-1\) (not including \(-1\)) towards the right end of the number line (towards \(\infty\)). Again, we use an open circle at \(-1\) for this part as well.

To graph it:

1. Locate the point \(-1\) on the number line.
2. Draw an open circle at \(-1\) (since \(y = - 1\) does not satisfy \(|y + 1|>0\), as \(| - 1+1|=|0| = 0\) which is not greater than 0).
3. Draw a line with an arrow pointing to the left from the open circle (representing \(y<-1\)) and a line with an arrow pointing to the right from the open circle (representing \(y>-1\)).

Answer:

The graph has an open circle at \(y=-1\), with a ray extending to the left of \(-1\) (towards \(-\infty\)) and a ray extending to the right of \(-1\) (towards \(\infty\)).