Question

Identify the square root function that is increasing over the interval [−3,∞).(1 point)
Responses

y
=
√
−
x
−
3
+
2
y is equal to square root of negative x minus 3 end root plus 2

y
=
√
x
−
3
+
2
y is equal to square root of x minus 3 end root plus 2

y
=
√
x
+
3
+
2
y is equal to square root of x plus 3 end root plus 2

y
=
−
√
x
+
3

Answer 1

To determine which square root function is increasing over the interval \([-3, \infty)\), we need to analyze each of the provided functions.

1. \(y = \sqrt{-x - 3} + 2\)
- The \( \sqrt{-x - 3} \) is defined when \(-x - 3 \geq 0\), or when \(x \leq -3\). This function is not increasing over the interval \([-3, \infty)\), as it is only defined for \(x\) values less than or equal to -3.

2. \(y = \sqrt{x - 3} + 2\)
- The \( \sqrt{x - 3} \) is defined when \(x - 3 \geq 0\), or when \(x \geq 3\). This means that it is not increasing on the interval \([-3, \infty)\) since it starts at \(x = 3\).

3. \(y = \sqrt{x + 3} + 2\)
- The \( \sqrt{x + 3} \) is defined for \(x + 3 \geq 0\), which simplifies to \(x \geq -3\). This function is defined on \([-3, \infty)\), and since the square root function is increasing, this function is increasing on this interval.

4. \(y = -\sqrt{x + 3}\)
- The \( -\sqrt{x + 3} \) is also defined for \(x + 3 \geq 0\), meaning \(x \geq -3\). However, this function is decreasing because of the negative sign in front of the square root.

Based on the analysis above, the function that is increasing over the interval \([-3, \infty)\) is:

**\(y = \sqrt{x + 3} + 2\)**.