Question

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

Answer 1

Graph opens down.
Vertex at `(0,-1)`;Axis of Symmetry: `x=0`

Explanation:

`y = -x^2-1`

`y` is a quadratic function of the from: `ax^2+bx+c`
Where: `a=-1, b=0 and c=-1`

Since `y` is a quadratic its graph will be a parabola.

Since `a<0` `y` will have a maximum value at its vertex.

Now, since `y` is a parabola with a maximum value it must open downwards.

The vertex of `y` will lie on its axis of symmetry where `x = (-b)/(2a)` i.e. where `x=0`

Since, `y(0) = -1 -> y_"vertex" = (0,-1)`

The axis of symmetry of `y` is the vertical line `x=0` which is the `y-`axis.

We can see these results from the graph of `y` below.
graph{ -x^2-1 [-13.28, 12.03, -7.17, 5.49]}