Question

Which function has a vertex at the origin? Help

Answer 1

Answer:

$f(x)=-x^2$

Step-by-step explanation:

Vertex form of a quadratic function:

  $y=a(x-h)^2+k$  

where:

• (h, k) is the vertex
• a is some constant
  If a > 0, the parabola opens upwards
  If a < 0, the parabola opens downwards

If the vertex is at the origin:

• h = 0
• k = 0

Substituting the vertex into the equation:

$\implies y=a(x-0)^2+0$

$\implies y= ax^2$

Comparing with the available answer options:

$f(x)=-x^2$  has its vertex at the origin.

Additional Information:

$\textsf{The vertex of }\:f(x)=(x+4)^2 \: \textsf{ is }\:(-4,0)$

$\textsf{The vertex of }\:f(x)=x(x-4) \: \textsf{ is }\:(2,-4)$

$\textsf{The vertex of }\:f(x)=(x-4)(x+4) \: \textsf{ is }\:(0,-16)$

Learn more about vertex form here:

brainly.com/question/27796555

Answer 2

Answer:

Option D

Step-by-step explanation:

• y=-x²

This function has vertex at origin

Let's verify

Put (0,0)

• 0=-(0)²
• 0=-0
• 0=0

Hence verified