To determine which function has its vertex at the origin, we should convert each function into its vertex form. The general vertex form of a quadratic function is [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 step by step:
1. [tex]\( f(x) = (x+4)^2 \)[/tex]
This function is already in the vertex form: [tex]\( f(x) = (x - (-4))^2 + 0 \)[/tex].
Hence, the vertex is [tex]\((-4, 0)\)[/tex], which does not lie at the origin.
2. [tex]\( f(x) = x(x-4) \)[/tex]
First, expand the function: [tex]\( f(x) = x^2 - 4x \)[/tex].
To convert this to vertex form, complete the square:
[tex]\[
f(x) = x^2 - 4x = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4
\][/tex]
So, the vertex form is [tex]\( f(x) = (x - 2)^2 - 4 \)[/tex].
Hence, the vertex is [tex]\((2, -4)\)[/tex], which does not lie at the origin.
3. [tex]\( f(x) = (x-4)(x+4) \)[/tex]
First, expand the function: [tex]\( f(x) = x^2 - 16 \)[/tex].
In vertex form this is written as [tex]\( f(x) = 1(x - 0)^2 - 16 \)[/tex].
Hence, the vertex is [tex]\((0, -16)\)[/tex], which does not lie at the origin.
4. [tex]\( f(x) = -x^2 \)[/tex]
This function is already in vertex form: [tex]\( f(x) = -1(x - 0)^2 + 0 \)[/tex].
So, the vertex is [tex]\((0, 0)\)[/tex], which lies at the origin.
Only the function [tex]\( f(x) = -x^2 \)[/tex] has its vertex at the origin [tex]\((0, 0)\)[/tex].
Therefore, the function that has a vertex at the origin is:
[tex]\[ f(x) = -x^2 \][/tex]
