Answer :
To determine which function has its vertex at the origin, we need to analyze each function in detail and find their respective vertices.
1. For the function \( f(x) = (x + 4)^2 \):
- This is a quadratic function in vertex form, which is generally written as \( f(x) = (x - h)^2 + k \), where \((h, k)\) is the vertex.
- For \( f(x) = (x + 4)^2 \), we can rewrite it as \( f(x) = (x - (-4))^2 \).
- Therefore, the vertex is at \( (-4, 0) \).
2. For the function \( f(x) = x(x - 4) \):
- First, expand the function to \( f(x) = x^2 - 4x \).
- This function is in the standard quadratic form \( ax^2 + bx + c \).
- The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
- Here, \( a = 1 \) and \( b = -4 \), so \( x = -\frac{-4}{2 \cdot 1} = 2 \).
- Substitute \( x = 2 \) back into the function to find the y-coordinate: \( f(2) = 2(2 - 4) = 2(-2) = -4 \).
- Therefore, the vertex is at \( (2, -4) \).
3. For the function \( f(x) = (x - 4)(x + 4) \):
- Expand the function to \( f(x) = x^2 - 16 \).
- This function is also in the standard quadratic form \( ax^2 + bx + c \) with \( a = 1 \), \( b = 0 \), and \( c = -16 \).
- The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
- Here, \( b = 0 \), so \( x = -\frac{0}{2 \cdot 1} = 0 \).
- Substitute \( x = 0 \) back into the function to find the y-coordinate: \( f(0) = (0 - 4)(0 + 4) = (-4)(4) = -16 \).
- Therefore, the vertex is at \( (0, -16) \).
4. For the function \( f(x) = -x^2 \):
- This function is also in the standard quadratic form \( ax^2 + bx + c \) with \( a = -1 \), \( b = 0 \), and \( c = 0 \).
- The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
- Here, \( b = 0 \), so \( x = -\frac{0}{2 \cdot (-1)} = 0 \).
- Substitute \( x = 0 \) back into the function to find the y-coordinate: \( f(0) = -0^2 = 0 \).
- Therefore, the vertex is at \( (0, 0) \).
Given this analysis, the function \( f(x) = -x^2 \) has its vertex at the origin \( (0,0) \).
The correct function with a vertex at the origin is:
[tex]\[ f(x) = -x^2 \][/tex]
1. For the function \( f(x) = (x + 4)^2 \):
- This is a quadratic function in vertex form, which is generally written as \( f(x) = (x - h)^2 + k \), where \((h, k)\) is the vertex.
- For \( f(x) = (x + 4)^2 \), we can rewrite it as \( f(x) = (x - (-4))^2 \).
- Therefore, the vertex is at \( (-4, 0) \).
2. For the function \( f(x) = x(x - 4) \):
- First, expand the function to \( f(x) = x^2 - 4x \).
- This function is in the standard quadratic form \( ax^2 + bx + c \).
- The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
- Here, \( a = 1 \) and \( b = -4 \), so \( x = -\frac{-4}{2 \cdot 1} = 2 \).
- Substitute \( x = 2 \) back into the function to find the y-coordinate: \( f(2) = 2(2 - 4) = 2(-2) = -4 \).
- Therefore, the vertex is at \( (2, -4) \).
3. For the function \( f(x) = (x - 4)(x + 4) \):
- Expand the function to \( f(x) = x^2 - 16 \).
- This function is also in the standard quadratic form \( ax^2 + bx + c \) with \( a = 1 \), \( b = 0 \), and \( c = -16 \).
- The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
- Here, \( b = 0 \), so \( x = -\frac{0}{2 \cdot 1} = 0 \).
- Substitute \( x = 0 \) back into the function to find the y-coordinate: \( f(0) = (0 - 4)(0 + 4) = (-4)(4) = -16 \).
- Therefore, the vertex is at \( (0, -16) \).
4. For the function \( f(x) = -x^2 \):
- This function is also in the standard quadratic form \( ax^2 + bx + c \) with \( a = -1 \), \( b = 0 \), and \( c = 0 \).
- The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
- Here, \( b = 0 \), so \( x = -\frac{0}{2 \cdot (-1)} = 0 \).
- Substitute \( x = 0 \) back into the function to find the y-coordinate: \( f(0) = -0^2 = 0 \).
- Therefore, the vertex is at \( (0, 0) \).
Given this analysis, the function \( f(x) = -x^2 \) has its vertex at the origin \( (0,0) \).
The correct function with a vertex at the origin is:
[tex]\[ f(x) = -x^2 \][/tex]
