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julia-pushkina [17]
1 year ago
11

Which function has a vertex at the origin? Help

Mathematics
2 answers:
I am Lyosha [343]1 year ago
8 0

Answer:

f(x)=-x^2

Step-by-step explanation:

<u>Vertex form of a quadratic function</u>:

  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 <u>vertex</u> is at the origin:

  • h = 0
  • k = 0

<u>Substituting</u> the vertex into the equation:

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

\implies y= ax^2

<u>Comparing</u> with the available answer options:

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

<u>Additional Information</u>:

\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

kari74 [83]1 year ago
5 0

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

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