Question

PLS HELP!!! 80PTS
I have attached a photo.

Answer 1

`y=-4x^2`

Explanation:

As the graph is a parabola, the equation is quadratic and so includes a term in x².

From inspection of the given graph:

• Vertex = (0, 0)
• Two points on the curve = (-1, -4) and (1, -4).

Vertex form of a quadratic equation:

`\boxed{y=a(x-h)^2+k}`

Where:

• a is a constant to be found.
• (h, k) is the vertex.

Substitute the vertex (0, 0) into the formula :

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

`\implies y=ax^2`

Substitute point (1, -4) into the equation and solve for a:

`\implies -4=a(1)^2`

`\implies -4=a`

`\implies a=-4`

Therefore, the equation of the parabola is:

`\boxed{y=-4x^2}`