Question

I need help finding 5 points. the vertex, 2 to the left of the vertex, and 2 points to the right of the vertex.

Answer 1

Let's convert the given equation first into a vertex form.

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

where (h, k) is the vertex.

The vertex form of the equatio that we have is:

`y=-2(x-0)^2+0`

Hence, the vertex of the equation is at the origin (0, 0).

Since "a" is negative, our parabola is opening downward.

Let's identify two points to the left of the vertex. Let's say at x = -1. Replace "x" with -1 in the equation.

`\begin{gathered} y=-2(-1)^2 \\ y=-2(1) \\ y=-2 \end{gathered}`

Hence, we have a point to the left of the parabola at (-1, -2).

Let's say x = -2. Replace "x" with -2 in the equation.

`\begin{gathered} y=-2(-2)^2 \\ y=-2(4) \\ y=-8 \end{gathered}`

Hence, we also have another point to the left of the parabola at (-2, -8).

If our x is to the right of the vertex, say, x = 1. Replace "x" with 1 in the equation.

`\begin{gathered} y=-2(1)^2 \\ y=-2(1) \\ y=-2 \end{gathered}`

We have a point to the right of the parabola at (1, -2).

If x = 2, let's replace "x" with 2 in the equation.

`\begin{gathered} y=-2(2)^2 \\ y=-2(4) \\ y=-8 \end{gathered}`

Hence, we also have another point to the right of the parabola at (2, -8).

The graph of this equation is: