Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given `y=-x^2-2x-1`?

Answer 1

The graph opens down.
The vertex is `(-1, 2)`.
The axis of symmetry is `x = -1`.

Explanation:

The graph opens down because the quadratic slope is negative. `(-x) ->` slope is `-1` .

This equation is in standard form, meaning that it has the form `ax^2 + bx + c`.
To find the vertex, we have to use `-b/(2a)` to find the x-coordinate of the vertex.

So we know a is -1 and b is -2. Let's put that into the formula `-b/(2a)`.
`(-(-2))/(2(-1))`

`2/-2`

`-1` The x-coordinate of the vertex is `-1`.

Now, in order to find the y-coordinate of the vertex, we we plug the x-coordinate value of the vertex back into the equation where x is:
`y_v = -(-1) - 2(-1) -1` (`y_v` means y value of vertex)
`y_v = 1 + 2 - 1`
`y_v = 2`

So the coordinate of the vertex is `(-1, 2)`.

Now, the axis of symmetry means where the graph reflects.
So we know that the vertex is `(-1, 2)`.

It becomes symmetrical at the line of `x = -1`, or the line of the x-axis at the vertex.