To determine which function has the range [tex]\(\{y \mid y \leq 5\}\)[/tex], we should analyze each of the given functions individually.
Each of these functions is a quadratic function in the form [tex]\(f(x) = a(x - h)^2 + k\)[/tex], where [tex]\(a\)[/tex], [tex]\(h\)[/tex], and [tex]\(k\)[/tex] are constants, and their graph is a parabola. The value of [tex]\(a\)[/tex] determines the direction the parabola opens (upwards if [tex]\(a > 0\)[/tex], and downwards if [tex]\(a < 0\)[/tex]), and the value of [tex]\(k\)[/tex] represents the vertex of the parabola.
1. [tex]\(f(x) = (x-4)^2 + 5\)[/tex]
- The coefficient [tex]\(a\)[/tex] is positive ([tex]\(a = 1 > 0\)[/tex]).
- The vertex is at [tex]\((4, 5)\)[/tex].
- The parabola opens upwards because [tex]\(a > 0\)[/tex].
- Therefore, the minimum value of [tex]\(f(x)\)[/tex] is 5, and the range is [tex]\(\{y \mid y \geq 5\}\)[/tex].
2. [tex]\(f(x) = -(x-4)^2 + 5\)[/tex]
- The coefficient [tex]\(a\)[/tex] is negative ([tex]\(a = -1 < 0\)[/tex]).
- The vertex is at [tex]\((4, 5)\)[/tex].
- The parabola opens downwards because [tex]\(a < 0\)[/tex].
- Therefore, the maximum value of [tex]\(f(x)\)[/tex] is 5, and the range is [tex]\(\{y \mid y \leq 5\}\)[/tex].
3. [tex]\(f(x) = (x-5)^2 + 4\)[/tex]
- The coefficient [tex]\(a\)[/tex] is positive ([tex]\(a = 1 > 0\)[/tex]).
- The vertex is at [tex]\((5, 4)\)[/tex].
- The parabola opens upwards because [tex]\(a > 0\)[/tex].
- Therefore, the minimum value of [tex]\(f(x)\)[/tex] is 4, and the range is [tex]\(\{y \mid y \geq 4\}\)[/tex].
4. [tex]\(f(x) = -(x-5)^2 + 4\)[/tex]
- The coefficient [tex]\(a\)[/tex] is negative ([tex]\(a = -1 < 0\)[/tex]).
- The vertex is at [tex]\((5, 4)\)[/tex].
- The parabola opens downwards because [tex]\(a < 0\)[/tex].
- Therefore, the maximum value of [tex]\(f(x)\)[/tex] is 4, and the range is [tex]\(\{y \mid y \leq 4\}\)[/tex].
After analyzing each function, we can conclude that the function [tex]\(f(x) = -(x-4)^2 + 5\)[/tex] has a range of [tex]\(\{y \mid y \leq 5\}\)[/tex].
