How do you graph the quadratic function and identify the vertex and axis of symmetry for #y=x^2-2x-1#?

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Answer 1 · 2018-02-12T17:16:22

Vertex: #(1,-2)#
Axis of Symmetry: #x=1#

You first convert to vertex form:

#y=a(x-h)^2+k# with #(h,k)# being the vertex. To get to this, you have to complete the square.

#y=(x-2x+1^2-1^2)-1#

#y=(x-1)^2-2#

Since the vertex is #(h,k)#, then the vertex here is #(1,-2)#.

The axis of symmetry is just the x-coordinate of the vertex or #x=-b/(2a)# in #y=ax^2+bx+c#:

#x=1# is the axis of symmetry.