What is the vertex form of #y=x^2 - 2 x - 5 #?

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Answer 1 · 2018-05-17T20:04:59

Vertex form is #y= (x-1)^2 -6#

The vertex is the point #(1,-6)#

Vertex form is #y = a(x+p)^2 +q# with the vertex at #(-p, q)#

This is derived by the process of completing the square.

The quadratic trinomial #x^2 -2x-5# is in the form #ax^2 +bx+c#

#x^2 -2x-5# is not a perfect square.

We need to add the correct constant which will make it a square.
This is found from #(b/2)^2# which in this case is #((-2)/2)^2 = color(blue)(1)#

#y = x^2 -2x color(blue)(+1) color(blue)(-1) -5" "larr (color(blue)(+1-1=0))#

#y = (x^2 -2x color(blue)(+1))+ ( color(blue)(-1) -5)#

#y= (x-1)^2 -6" "larr# this is vertex form.

The vertex is the point #(1,-6)#