Asked by Jesusislord!❤️
The graph of which function is decreasing over the interval (–4, ∞)?
f(x) = (x + 4)2 + 4
f(x) = –(x + 4)2 + 4
f(x) = (x – 4)2 – 4
f(x) = –(x – 4)2 – 4
f(x) = (x + 4)2 + 4
f(x) = –(x + 4)2 + 4
f(x) = (x – 4)2 – 4
f(x) = –(x – 4)2 – 4
Answers
Answered by
GPT-4o mini
To determine which function is decreasing over the interval \((-4, \infty)\), we can analyze each of the given functions.
1. **For** \(f(x) = (x + 4)^2 + 4\):
- This is a quadratic function that opens upwards (since the coefficient of \(x^2\) is positive).
- The vertex is at \((-4, 4)\), and the function is decreasing on \((-4, -4]\) and increasing on \([-4, \infty)\). Therefore, it's not decreasing over the entire interval \((-4, \infty)\).
2. **For** \(f(x) = -(x + 4)^2 + 4\):
- This is a quadratic function that opens downwards (since the coefficient of \(x^2\) is negative).
- The vertex is at \((-4, 4)\), and the function is increasing on \((-\infty, -4]\) and decreasing on \([-4, \infty)\). Therefore, it is decreasing over the interval \((-4, \infty)\).
3. **For** \(f(x) = (x - 4)^2 - 4\):
- This is another quadratic function that opens upwards.
- The vertex is at \((4, -4)\). It is decreasing on \((- \infty, 4]\) and increasing on \([4, \infty)\), thus it is not decreasing over the whole interval \((-4, \infty)\).
4. **For** \(f(x) = -(x - 4)^2 - 4\):
- This is a quadratic function that opens downwards.
- The vertex is at \((4, -4)\), and it is increasing on \((-\infty, 4]\) and decreasing on \([4, \infty)\). Thus, it is also not decreasing over the entire interval \((-4, \infty)\).
**Conclusion**: The function that is decreasing over the interval \((-4, \infty)\) is:
\[
\boxed{f(x) = -(x + 4)^2 + 4}
\]
1. **For** \(f(x) = (x + 4)^2 + 4\):
- This is a quadratic function that opens upwards (since the coefficient of \(x^2\) is positive).
- The vertex is at \((-4, 4)\), and the function is decreasing on \((-4, -4]\) and increasing on \([-4, \infty)\). Therefore, it's not decreasing over the entire interval \((-4, \infty)\).
2. **For** \(f(x) = -(x + 4)^2 + 4\):
- This is a quadratic function that opens downwards (since the coefficient of \(x^2\) is negative).
- The vertex is at \((-4, 4)\), and the function is increasing on \((-\infty, -4]\) and decreasing on \([-4, \infty)\). Therefore, it is decreasing over the interval \((-4, \infty)\).
3. **For** \(f(x) = (x - 4)^2 - 4\):
- This is another quadratic function that opens upwards.
- The vertex is at \((4, -4)\). It is decreasing on \((- \infty, 4]\) and increasing on \([4, \infty)\), thus it is not decreasing over the whole interval \((-4, \infty)\).
4. **For** \(f(x) = -(x - 4)^2 - 4\):
- This is a quadratic function that opens downwards.
- The vertex is at \((4, -4)\), and it is increasing on \((-\infty, 4]\) and decreasing on \([4, \infty)\). Thus, it is also not decreasing over the entire interval \((-4, \infty)\).
**Conclusion**: The function that is decreasing over the interval \((-4, \infty)\) is:
\[
\boxed{f(x) = -(x + 4)^2 + 4}
\]
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