To determine which function has its vertex at the origin [tex]\((0,0)\)[/tex], we analyze the vertex of each given quadratic function.
1. Function: [tex]\( f(x) = (x+4)^2 \)[/tex]
- This is in vertex form, [tex]\( f(x) = (x-h)^2 + k \)[/tex], where the vertex is located at [tex]\((h, k)\)[/tex].
- Here, [tex]\( h = -4 \)[/tex] and [tex]\( k = 0 \)[/tex], so the vertex is [tex]\((-4, 0)\)[/tex].
2. Function: [tex]\( f(x) = x(x-4) \)[/tex]
- Expand the function: [tex]\( f(x) = x^2 - 4x \)[/tex].
- For a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex], the vertex occurs at [tex]\( x = -\frac{b}{2a} \)[/tex].
- Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -4 \)[/tex], so [tex]\( x = -\frac{-4}{2 \cdot 1} = 2 \)[/tex].
- Substitute [tex]\( x = 2 \)[/tex] back into the function: [tex]\( f(2) = 2^2 - 4 \cdot 2 = 4 - 8 = -4 \)[/tex].
- The vertex is [tex]\( (2, -4) \)[/tex].
3. Function: [tex]\( f(x) = (x-4)(x+4) \)[/tex]
- Expand the function: [tex]\( f(x) = x^2 - 16 \)[/tex].
- This is in standard form [tex]\( f(x) = ax^2 + bx + c \)[/tex] with [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = -16 \)[/tex].
- The vertex occurs at [tex]\( x = -\frac{b}{2a} = 0 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] back into the function: [tex]\( f(0) = 0^2 - 16 = -16 \)[/tex].
- The vertex is [tex]\( (0, -16) \)[/tex].
4. Function: [tex]\( f(x) = -x^2 \)[/tex]
- This is a downward-opening parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
- The vertex form of this quadratic function is already given by [tex]\( f(x) = -x^2 + 0 \)[/tex].
- Here, the vertex is at [tex]\( (0, 0) \)[/tex].
Among the given functions, the one with the vertex at the origin, [tex]\((0, 0)\)[/tex], is:
[tex]\[ f(x) = -x^2 \][/tex]
Thus, the correct answer is:
4. [tex]\( f(x) = -x^2 \)[/tex]
