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#?

1 Answer
Oct 26, 2017

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.