Question

(-2,-4)

Oy=(x + 2)² +4
Oy=(x - 2)²-4
Oy=-(x - 2)²-4
Oy=-(x + 2)² - 4

Answer 1

Answer:

$\textsf{D)} \quad y=-(x+2)^2-4$

Step-by-step explanation:

Vertex form of a quadratic equation:  

$y=a(x-h)^2+k$

where:

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

From inspection of the graph, the vertex (turning point) is (-2, -4).

Substituting the vertex into the equation:

$\implies y=a(x-(-2))^2+(-4)$

$\implies y=a(x+2)^2-4$

As the parabola opens downwards, $a < 0$:

$\implies y=-a(x+2)^2-4$

The curve intercepts the y-axis at (0, -8).

Inputting this into the equation and solving for $a$:

$\implies -8=-a(0+2)^2-4$

$\implies -4=-4a$

$\implies a=1$

Therefore, the equation of the graph is:

$y=-(x+2)^2-4$

Answer 2

Answer:

y = -(x + 2)² - 4

Step-by-step explanation:

Vertex form of an equation :

y = a(x - h)² + k

• Vertex = (h, k)

Substituting the coordinates of the vertex in the equation :

y = (x -(-2))² + (-4)

y = (x + 2)² - 4

y = -(x + 2)² - 4 (as the graph  opens downwards)