To solve the inequality [tex]\( y^2 \geq 9 \)[/tex], follow these steps:
1. Rewrite the inequality:
[tex]\[
y^2 - 9 \geq 0
\][/tex]
2. Factor the quadratic expression:
[tex]\[
y^2 - 9 = (y - 3)(y + 3)
\][/tex]
So, the inequality becomes:
[tex]\[
(y - 3)(y + 3) \geq 0
\][/tex]
3. Find the critical points by setting each factor equal to zero:
[tex]\[
y - 3 = 0 \quad \Rightarrow \quad y = 3
\][/tex]
[tex]\[
y + 3 = 0 \quad \Rightarrow \quad y = -3
\][/tex]
These critical points divide the number line into three intervals:
[tex]\[
(-\infty, -3) \quad , \quad [-3, 3] \quad , \quad (3, \infty)
\][/tex]
4. Determine the sign of the expression [tex]\((y - 3)(y + 3)\)[/tex] in each interval:
- For [tex]\( y \in (-\infty, -3) \)[/tex]:
Choose [tex]\( y = -4 \)[/tex]:
[tex]\[
(y - 3)(y + 3) = (-4 - 3)(-4 + 3) = (-7)(-1) = 7 \geq 0
\][/tex]
So, the inequality holds.
- For [tex]\( y \in (-3, 3) \)[/tex]:
Choose [tex]\( y = 0 \)[/tex]:
[tex]\[
(y - 3)(y + 3) = (0 - 3)(0 + 3) = (-3)(3) = -9 < 0
\][/tex]
So, the inequality does not hold.
- For [tex]\( y \in (3, \infty) \)[/tex]:
Choose [tex]\( y = 4 \)[/tex]:
[tex]\[
(y - 3)(y + 3) = (4 - 3)(4 + 3) = (1)(7) = 7 \geq 0
\][/tex]
So, the inequality holds.
5. Check the endpoints [tex]\( y = -3 \)[/tex] and [tex]\( y = 3 \)[/tex]:
[tex]\[
(y - 3)(y + 3) = 0 \quad \text{at} \quad y = -3 \quad \text{and} \quad y = 3
\][/tex]
Since 0 is included in [tex]\(\geq 0\)[/tex], both endpoints are included.
6. Combine the intervals where the inequality holds:
[tex]\[
y \in (-\infty, -3] \cup [3, \infty)
\][/tex]
So, the solution set in interval notation is:
[tex]\[
(-\infty, -3] \cup [3, \infty)
\][/tex]
