Answer :
To determine which of the given functions has a vertex at [tex]\((2, -9)\)[/tex], we'll analyze the vertex form of each function. The vertex form of a quadratic function is given by:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
Where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Let's analyze each function individually:
1. Function: [tex]\( f(x) = -(x - 3)^2 \)[/tex]
- This is already in vertex form.
- Here, [tex]\(a = -1\)[/tex], [tex]\(h = 3\)[/tex], [tex]\(k = 0\)[/tex].
- The vertex of this function is [tex]\((3, 0)\)[/tex].
2. Function: [tex]\( f(x) = (x + 8)^2 \)[/tex]
- This is already in vertex form.
- Here, [tex]\(a = 1\)[/tex], [tex]\(h = -8\)[/tex], [tex]\(k = 0\)[/tex].
- The vertex of this function is [tex]\((-8, 0)\)[/tex].
3. Function: [tex]\( f(x) = (x - 5)(x + 1) \)[/tex]
- First, we need to expand this to find the standard form [tex]\(ax^2 + bx + c\)[/tex]:
[tex]\[ f(x) = x^2 + x - 5x - 5 = x^2 - 4x - 5 \][/tex]
- To find the vertex of this quadratic function, we use the formula for the vertex of a parabola given by a quadratic equation [tex]\(ax^2 + bx + c\)[/tex]:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex] and [tex]\(b = -4\)[/tex]:
[tex]\[ x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2 \][/tex]
- Substitute [tex]\(x = 2\)[/tex] back into the function to find the corresponding [tex]\(y\)[/tex]-coordinate:
[tex]\[ f(2) = (2 - 5)(2 + 1) = (-3)(3) = -9 \][/tex]
- The vertex of this function is [tex]\((2, -9)\)[/tex].
4. Function: [tex]\( f(x) = -(x - 1)(x - 5) \)[/tex]
- First, expand this to find the standard form:
[tex]\[ f(x) = -(x^2 - 6x + 5) = -x^2 + 6x - 5 \][/tex]
- To find the vertex of this quadratic function, we use the vertex formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\(a = -1\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[ x = -\frac{6}{2 \times (-1)} = -\frac{6}{-2} = 3 \][/tex]
- Substitute [tex]\(x = 3\)[/tex] back into the function to find the corresponding [tex]\(y\)[/tex]-coordinate:
[tex]\[ f(3) = -(3 - 1)(3 - 5) = -(2)(-2) = 4 \][/tex]
- The vertex of this function is [tex]\((3, 4)\)[/tex].
After analyzing all the functions, the function [tex]\( f(x) = (x - 5)(x + 1) \)[/tex] has the vertex at [tex]\((2, -9)\)[/tex]. Therefore, the correct function is:
[tex]\[ \boxed{f(x) = (x - 5)(x + 1)} \][/tex]
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
Where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Let's analyze each function individually:
1. Function: [tex]\( f(x) = -(x - 3)^2 \)[/tex]
- This is already in vertex form.
- Here, [tex]\(a = -1\)[/tex], [tex]\(h = 3\)[/tex], [tex]\(k = 0\)[/tex].
- The vertex of this function is [tex]\((3, 0)\)[/tex].
2. Function: [tex]\( f(x) = (x + 8)^2 \)[/tex]
- This is already in vertex form.
- Here, [tex]\(a = 1\)[/tex], [tex]\(h = -8\)[/tex], [tex]\(k = 0\)[/tex].
- The vertex of this function is [tex]\((-8, 0)\)[/tex].
3. Function: [tex]\( f(x) = (x - 5)(x + 1) \)[/tex]
- First, we need to expand this to find the standard form [tex]\(ax^2 + bx + c\)[/tex]:
[tex]\[ f(x) = x^2 + x - 5x - 5 = x^2 - 4x - 5 \][/tex]
- To find the vertex of this quadratic function, we use the formula for the vertex of a parabola given by a quadratic equation [tex]\(ax^2 + bx + c\)[/tex]:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex] and [tex]\(b = -4\)[/tex]:
[tex]\[ x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2 \][/tex]
- Substitute [tex]\(x = 2\)[/tex] back into the function to find the corresponding [tex]\(y\)[/tex]-coordinate:
[tex]\[ f(2) = (2 - 5)(2 + 1) = (-3)(3) = -9 \][/tex]
- The vertex of this function is [tex]\((2, -9)\)[/tex].
4. Function: [tex]\( f(x) = -(x - 1)(x - 5) \)[/tex]
- First, expand this to find the standard form:
[tex]\[ f(x) = -(x^2 - 6x + 5) = -x^2 + 6x - 5 \][/tex]
- To find the vertex of this quadratic function, we use the vertex formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\(a = -1\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[ x = -\frac{6}{2 \times (-1)} = -\frac{6}{-2} = 3 \][/tex]
- Substitute [tex]\(x = 3\)[/tex] back into the function to find the corresponding [tex]\(y\)[/tex]-coordinate:
[tex]\[ f(3) = -(3 - 1)(3 - 5) = -(2)(-2) = 4 \][/tex]
- The vertex of this function is [tex]\((3, 4)\)[/tex].
After analyzing all the functions, the function [tex]\( f(x) = (x - 5)(x + 1) \)[/tex] has the vertex at [tex]\((2, -9)\)[/tex]. Therefore, the correct function is:
[tex]\[ \boxed{f(x) = (x - 5)(x + 1)} \][/tex]
