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#?

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Answer 1 · 2018-06-16T03:52:07

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

#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]}