What is the vertex of #y= x^2 - x - 6 #?

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Answer 1 · 2017-04-29T15:19:00

Vertex is #(0.5, -6.25)#

#y = x^2 -x-6#

Find the axis of symmetry first using the formula:

#x = (-b)/(2a)" rarr x = (-(-1))/(2*1)#

#x = 1/2#

Then find #y# by substituting #x = 1/2#

#y = (1/2)^2 -1/2 -6#

#y= -6 1/4#

Vertex is #(0.5, -6.25)#