Question

Identify the square root function that has a range of (−∞,−5].(1 point)
Responses

y
=
√
−
x
−
5
y is equal to square root of negative x minus 5

y
=
−
√
x
−
3
+
5
y is equal to negative square root of x minus 3 end root plus 5

y
=
√
x
−
3
−
5
y is equal to square root of x minus 3 end root minus 5

y
=
−
√
−
x
−
5

Answer 1

To find a square root function with a range of \((-\infty, -5]\), we need to analyze the options given.

1. **\(y = \sqrt{-x} - 5\)**
- The square root function outputs values from \(0\) to \(\infty\). The transformation \(y = \sqrt{-x}\) puts the outputs in the range \([0, \infty]\).
- Subtracting \(5\) shifts this range to \([-5, \infty]\), which does not match our requirements.

2. **\(y = -\sqrt{x - 3} + 5\)**
- The square root function \(\sqrt{x - 3}\) has a range of \([0, \infty]\) since it starts from \(x = 3\).
- By multiplying by \(-1\) and adding \(5\), the range transforms to \([-5, \infty]\). Thus, this function does not meet our requirement.

3. **\(y = \sqrt{x - 3} - 5\)**
- Similar to previous examples, \(\sqrt{x - 3}\) has a range of \([0, \infty]\), so subtracting \(5\) gives a range of \([-5, \infty]\), which again does not fit.

4. **\(y = -\sqrt{-x} - 5\)**
- The square root function \(\sqrt{-x}\) transforms the inputs where \(x \leq 0\) to ranges starting at \(0\). Therefore, \(-\sqrt{-x}\) will have a range of \((-\infty, 0]\).
- Then, subtracting \(5\) changes the range to \((-\infty, -5]\), which fits our requirement.

Therefore, the correct function that has a range of \((-\infty, -5]\) is:

**\(y = -\sqrt{-x} - 5\)**.