Question

which function has a vertex at the origin?
f(x) = (x + 4)^2
f(x) = x(x - 4)
f(x) = (x - 4)(x + 4)
f(x) = -x^2

Explanation:

Step1: Recall vertex form of quadratic

The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For the vertex to be at the origin \((0, 0)\), \(h = 0\) and \(k = 0\).

Step2: Analyze each function

• For \( f(x) = (x + 4)^2 \), expand it: \( f(x) = (x + 4)^2 = (x - (-4))^2 + 0 \). The vertex is \((-4, 0)\), not the origin.
• For \( f(x) = x(x - 4) = x^2 - 4x \). Complete the square: \( x^2 - 4x = (x - 2)^2 - 4 \). Vertex is \((2, -4)\), not the origin.
• For \( f(x) = (x - 4)(x + 4) = x^2 - 16 \). This is \( f(x) = (x - 0)^2 - 16 \), so vertex is \((0, -16)\), not the origin.
• For \( f(x) = -x^2 \), rewrite in vertex form: \( f(x) = -1(x - 0)^2 + 0 \). Here, \( h = 0 \) and \( k = 0 \), so vertex is \((0, 0)\) (the origin).

Answer:

D. \( f(x) = -x^2 \) (assuming the options are labeled A to D as \( f(x) = (x + 4)^2 \), \( f(x) = x(x - 4) \), \( f(x) = (x - 4)(x + 4) \), \( f(x) = -x^2 \) respectively)