Question

Which function would change the range of p(x)=−3x+2−−−−√+7
𝑝
(
𝑥
)
=
−
3
𝑥
+
2
+
7
to {y|y≤−1}
{
𝑦
|
𝑦
≤
−
1
}
?(1 point)
Responses

d(x)=p(x)−1
𝑑
(
𝑥
)
=
𝑝
(
𝑥
)
−
1
d left parenthesis x right parenthesis equals p left parenthesis x right parenthesis minus 1

w(x)=−p(x)−8
𝑤
(
𝑥
)
=
−
𝑝
(
𝑥
)
−
8
w left parenthesis x right parenthesis equals negative p left parenthesis x right parenthesis minus 8

f(x)=p(x)−8
𝑓
(
𝑥
)
=
𝑝
(
𝑥
)
−
8
f left parenthesis x right parenthesis equals p left parenthesis x right parenthesis minus 8

w(x)=p(x+1)

Answer 1

To change the range of the function \( p(x) = -3\sqrt{x} + 7 \) to \( \{y | y \leq -1\} \), we need to analyze the current range of \( p(x) \).

1. **Finding the range of \( p(x) \)**:
- The term \( -3\sqrt{x} \) will be non-positive for all \( x \geq 0 \) (since the square root is always non-negative and is multiplied by -3).
- At \( x = 0 \), \( p(0) = -3\sqrt{0} + 7 = 7 \).
- As \( x \) increases, \( -3\sqrt{x} \) decreases, meaning \( p(x) \) decreases.
- Therefore, the maximum value occurs at \( x = 0 \), and as \( x \) approaches infinity, \( p(x) \) approaches \( -\infty \).

The range of \( p(x) \) is \( (-\infty, 7] \).

2. **Changing the range to \( \{y | y \leq -1\} \)**:
- We want to adjust \( p(x) \) so that it only outputs values less than or equal to -1.
- The current maximum of \( p(x) \) is 7, and we need to shift all outputs down by a certain amount.

To adjust the function's output down to the desired range, we can apply a transformation:

- To shift the output down by 8, we can use the function \( f(x) = p(x) - 8 \).

This will change the maximum of \( p(x) \) (which is 7) to:

\[
7 - 8 = -1
\]

Thus, the new range will become \( (-\infty, -1] \), which is what we want.

### Final Answer
The function that changes the range of \( p(x) \) to \( \{y | y \leq -1\} \) is:

\[
f(x) = p(x) - 8
\]

This corresponds to the option:
\( f(x) = p(x) - 8 \)