To determine which quadratic function has a range described by \( y \leq 10.5 \), we need to consider the general shape and properties of the given quadratic functions.
Quadratic functions can be in standard form \( f(x) = ax^2 + bx + c \) or vertex form \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. The sign of \( a \) determines whether the parabola opens upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)).
For a range of \( y \leq 10.5 \), we need:
- The parabola must open downwards (which means \( a < 0 \)).
- The maximum value of the function must be equal to 10.5. In vertex form, this occurs at \( k = 10.5 \) when the vertex is at the maximum point.
Now let's analyze each of the provided options:
a. \( f(x) = 6(x + 4)^2 + 10.5 \)
- Here, \( a = 6 \) which is positive. Therefore, the parabola opens upwards. (Range is \( y \geq 10.5 \))
b. \( f(x) = 10.5(x - 4)^2 - 6 \)
- \( a = 10.5 \) which is also positive. Therefore, the parabola opens upwards. (Range is \( y \geq -6 \))
c. \( f(x) = -6(x - 10.5)^2 + 4 \)
- \( a = -6 \) which is negative. The parabola opens downwards. The vertex is at \( (10.5, 4) \), so the maximum value is 4, which does not satisfy \( y \leq 10.5 \). (Range is \( y \leq 4 \))
d. \( f(x) = -6(x + 4)^2 + 10.5 \)
- \( a = -6 \) which is negative. The parabola opens downwards. The vertex is at \( (-4, 10.5) \), so the maximum value is 10.5. This matches the criteria \( y \leq 10.5 \).
Given this analysis, the quadratic function that has a range described as \( y \leq 10.5 \) is:
d. \( f(x) = -6(x + 4)^2 + 10.5 \)
3 answers