# 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?

Jun 16, 2018

Graph opens down.
Vertex at $\left(0 , - 1\right)$;Axis of Symmetry: $x = 0$

#### Explanation:

$y = - {x}^{2} - 1$

$y$ is a quadratic function of the from: $a {x}^{2} + b x + c$
Where: $a = - 1 , b = 0 \mathmr{and} c = - 1$

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 $x = \frac{- b}{2 a}$ i.e. where $x = 0$

Since, $y \left(0\right) = - 1 \to {y}_{\text{vertex}} = \left(0 , - 1\right)$

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]}