Question
Which function would change the range of p(x)=−3x+2−−−−√+7
to {y|y≤−1}
?(1 point)
Responses
d(x)=p(x)−1
d left parenthesis x right parenthesis equals p left parenthesis x right parenthesis minus 1
w(x)=p(x+1)
w left parenthesis x right parenthesis equals p left parenthesis x plus 1 right parenthesis
f(x)=p(x)−8
f left parenthesis x right parenthesis equals p left parenthesis x right parenthesis minus 8
w(x)=−p(x)−8
to {y|y≤−1}
?(1 point)
Responses
d(x)=p(x)−1
d left parenthesis x right parenthesis equals p left parenthesis x right parenthesis minus 1
w(x)=p(x+1)
w left parenthesis x right parenthesis equals p left parenthesis x plus 1 right parenthesis
f(x)=p(x)−8
f left parenthesis x right parenthesis equals p left parenthesis x right parenthesis minus 8
w(x)=−p(x)−8
Answers
Answered by
GPT-4o mini
To change the range of the function \( p(x) = -3\sqrt{x} + 7 \) to \( \{y | y \leq -1\} \), we need to analyze the existing range of \( p(x) \).
First, we find the range of \( p(x) \):
1. The term \( -3\sqrt{x} \) decreases as \( x \) increases since the square root is always non-negative, making \( p(x) \) a decreasing function.
2. When \( x = 0 \), \( p(0) = 7 \).
3. As \( x \) approaches infinity, \( -3\sqrt{x} \) approaches negative infinity, so \( p(x) \) also approaches negative infinity.
Thus, the range of \( p(x) \) is \( (-\infty, 7] \).
To change the range to \( \{y | y \leq -1\} \), we need to shift and reflect the graph of \( p(x) \) appropriately.
1. If we subtract a constant, it shifts the entire graph down.
2. If we reflect it across the x-axis (for example, multiplying by -1), this will invert the outputs.
Looking through the provided options:
1. **\( d(x) = p(x) - 1 \)**: This shifts the range down to \( (-\infty, 6] \), still higher than -1.
2. **\( w(x) = p(x+1) \)**: This does not change the range significantly.
3. **\( f(x) = p(x) - 8 \)**: This shifts the range to \( (-\infty, -1] \), which matches our requirement.
4. **\( w(x) = -p(x) - 8 \)**: This reflects and shifts down but results in positive outputs, which is not suitable.
**Therefore, the function that changes the range of \( p(x) \) to \( \{y | y \leq -1\}\) is:**
**\( f(x) = p(x) - 8 \)**.
First, we find the range of \( p(x) \):
1. The term \( -3\sqrt{x} \) decreases as \( x \) increases since the square root is always non-negative, making \( p(x) \) a decreasing function.
2. When \( x = 0 \), \( p(0) = 7 \).
3. As \( x \) approaches infinity, \( -3\sqrt{x} \) approaches negative infinity, so \( p(x) \) also approaches negative infinity.
Thus, the range of \( p(x) \) is \( (-\infty, 7] \).
To change the range to \( \{y | y \leq -1\} \), we need to shift and reflect the graph of \( p(x) \) appropriately.
1. If we subtract a constant, it shifts the entire graph down.
2. If we reflect it across the x-axis (for example, multiplying by -1), this will invert the outputs.
Looking through the provided options:
1. **\( d(x) = p(x) - 1 \)**: This shifts the range down to \( (-\infty, 6] \), still higher than -1.
2. **\( w(x) = p(x+1) \)**: This does not change the range significantly.
3. **\( f(x) = p(x) - 8 \)**: This shifts the range to \( (-\infty, -1] \), which matches our requirement.
4. **\( w(x) = -p(x) - 8 \)**: This reflects and shifts down but results in positive outputs, which is not suitable.
**Therefore, the function that changes the range of \( p(x) \) to \( \{y | y \leq -1\}\) is:**
**\( f(x) = p(x) - 8 \)**.