Answer :
To determine which function has a vertex at the origin, we need to analyze each function individually.
1. Function: \( f(x) = (x+4)^2 \)
To find the vertex of this function, we can recognize that it is already in the vertex form:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
Here, \( a = 1 \), \( h = -4 \), and \( k = 0 \).
The vertex of this function is at \( (h, k) = (-4, 0) \).
Hence, the vertex is at \((-4, 0)\), not at the origin \((0, 0)\).
2. Function: \( f(x) = x(x-4) \)
To find the vertex, we rewrite the equation in standard form (i.e., \( f(x) = ax^2 + bx + c \)).
[tex]\[ f(x) = x^2 - 4x \][/tex]
The vertex of a quadratic function in the form \( ax^2 + bx + c \) is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, \( a = 1 \) and \( b = -4 \):
[tex]\[ x = -\frac{-4}{2 \cdot 1} = 2 \][/tex]
Plugging \( x = 2 \) back into the function to find \( y \):
[tex]\[ f(2) = 2(2-4) = 2(-2) = -4 \][/tex]
The vertex of this function is at \( (2, -4) \), not at the origin \((0, 0)\).
3. Function: \( f(x) = (x-4)(x+4) \)
First, we expand this function to its standard form.
[tex]\[ f(x) = x^2 - 16 \][/tex]
This function is now in the form \( f(x) = ax^2 + bx + c \) with \( a = 1 \), \( b = 0 \), and \( c = -16 \).
The vertex of a quadratic function \( ax^2 + bx + c \) is given by the same vertex formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Since \( b = 0 \):
[tex]\[ x = 0 \][/tex]
Plugging \( x = 0 \) back into the function to find \( y \):
[tex]\[ f(0) = 0^2 - 16 = -16 \][/tex]
The vertex of this function is at \( (0, -16) \), not at the origin \((0, 0)\).
4. Function: \( f(x) = -x^2 \)
This function is already in the vertex form:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
Here, \( a = -1 \), \( h = 0 \), and \( k = 0 \).
The vertex of this function is at \( (h, k) = (0, 0) \).
Hence, the vertex is at the origin \((0, 0)\).
Among the given functions, the function \( f(x) = -x^2 \) has a vertex at the origin. Therefore, the correct function is:
[tex]\[ f(x) = -x^2 \][/tex]
1. Function: \( f(x) = (x+4)^2 \)
To find the vertex of this function, we can recognize that it is already in the vertex form:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
Here, \( a = 1 \), \( h = -4 \), and \( k = 0 \).
The vertex of this function is at \( (h, k) = (-4, 0) \).
Hence, the vertex is at \((-4, 0)\), not at the origin \((0, 0)\).
2. Function: \( f(x) = x(x-4) \)
To find the vertex, we rewrite the equation in standard form (i.e., \( f(x) = ax^2 + bx + c \)).
[tex]\[ f(x) = x^2 - 4x \][/tex]
The vertex of a quadratic function in the form \( ax^2 + bx + c \) is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, \( a = 1 \) and \( b = -4 \):
[tex]\[ x = -\frac{-4}{2 \cdot 1} = 2 \][/tex]
Plugging \( x = 2 \) back into the function to find \( y \):
[tex]\[ f(2) = 2(2-4) = 2(-2) = -4 \][/tex]
The vertex of this function is at \( (2, -4) \), not at the origin \((0, 0)\).
3. Function: \( f(x) = (x-4)(x+4) \)
First, we expand this function to its standard form.
[tex]\[ f(x) = x^2 - 16 \][/tex]
This function is now in the form \( f(x) = ax^2 + bx + c \) with \( a = 1 \), \( b = 0 \), and \( c = -16 \).
The vertex of a quadratic function \( ax^2 + bx + c \) is given by the same vertex formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Since \( b = 0 \):
[tex]\[ x = 0 \][/tex]
Plugging \( x = 0 \) back into the function to find \( y \):
[tex]\[ f(0) = 0^2 - 16 = -16 \][/tex]
The vertex of this function is at \( (0, -16) \), not at the origin \((0, 0)\).
4. Function: \( f(x) = -x^2 \)
This function is already in the vertex form:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
Here, \( a = -1 \), \( h = 0 \), and \( k = 0 \).
The vertex of this function is at \( (h, k) = (0, 0) \).
Hence, the vertex is at the origin \((0, 0)\).
Among the given functions, the function \( f(x) = -x^2 \) has a vertex at the origin. Therefore, the correct function is:
[tex]\[ f(x) = -x^2 \][/tex]
