How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y = -x^2-2x+6#?

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Answer 1 · 2015-06-06T22:51:37

#y = -x^2-2x+6#

#= -(x^2+2x-6)#

#= -((x+1)^2-7)#

This has vertex where #(x+1)^2 = 0#, that is where #x = -1#
and #y = - (-7) = 7#. That is #(-1, 7)#

The axis of symmetry is the vertical line with equation #x = -1#

The intercept with the #y# axis is where #x=0# and #y=6#, that is at #(0, 6)#.

The intercepts with the #x# axis are the roots of

#-((x+1)^2-7) = 0#

Hence #(x+1)^2 = 7#

So

#x+1 = +-sqrt(7)#

#x = -1+-sqrt(7)#

So the intercepts are:

#(-1+sqrt(7), 0)# and #(-1-sqrt(7), 0)#