Respuesta :

Answer:

The correct option is y = -x² + 2

Step-by-step explanation:

Given:

  • parabola opens downward
  • extreme point at (0,2)

Parabola opens downward → Coefficient of x² has to be negative

Therefore, the 2nd and the 4th option are wrong.

Extreme point at (0, 2) → for this question, the fastest way is to substitute x and y with 0 and 2

  • 1st option: y = -2x² - 1

                         2 = -2(0)² - 1

                         2 = -1

        therefore, 1st option is incorrect

  • 3st option: y = -x² + 2
  •                          2 = -(0)² + 2
  •                          2 = 2
  •         therefore, 3rd option is incorrect

Answer:

[tex]\textsf{C)}\quad y = -x^2+2[/tex]

Step-by-step explanation:

The provided graph shows a downward-opening parabola with the vertex at (0, 2).

A downward-opening parabola is characterized by a negative coefficient for the squared term. Among the provided options, the equations with a negative coefficient for x² are:

[tex]y= -2x^2-1[/tex]

[tex]y = -x^2+2[/tex]

As the graphed parabola crosses the y-axis at (0, 2), the correct equation that represents the graph is y = -x² + 2. When x = 0 is substituted into this equation, y = 2, corresponding to the y-intercept (0, 2).

So, the equation that is represented by the provided graph is:

[tex]\Large\boxed{\boxed{y = -x^2+2}}[/tex]

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