Which is the vertex of #x^2+10x=-17#?

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Which is the vertex of #x^2+10x=-17#?

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Answer 1 · 2016-11-21T17:50:32

#(-5,-8)#

Explanation:

#x^2+10x=-17#
#0=-x^2-10x-17#
#0=-[x^2+10x+17]#
#0=-[(x+5)^2-8]#
#0=-(x+5)^2+8#

#x^2+10x+17=0#
#(x+5)^2-8=0#

Vertex is at #x = -5#. It is unclear whether the coefficient of the highest degree is positive or negative. If the parabola is negative, then the vertex is at #(-5,8)#. If the parabola is positive, the vertex is at #(-5,-8)#

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Which is the vertex of #x^2+10x=-17#?

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