What is the vertex of #y= 2(x - 4)^2 - x^2+4x-1 #?

1 Answer
Dec 16, 2015

vertex#=(6,-5)#

Explanation:

Start by expanding the brackets, then simplifying the terms:

#y=2(x-4)^2-x^2+4x-1#
#y=2(x-4)(x-4)-x^2+4x-1#
#y=2(x^2-8x+16)-x^2+4x-1#
#y=2x^2-16+32-x^2+4x-1#
#y=x^2-12x+31#

Take the simplified equation and rewrite it in vertex form:

#y=x^2-12x+31#
#y=(x^2-12x)+31#
#y=(x^2-12x+(12/2)^2-(12/2)^2)+31#
#y=(x^2-12x+(6)^2-(6)^2)+31#
#y=(x^2-12x+36-36)+31#
#y=(x^2-12x+36)+31-(36*1)#
#y=(x-6)^2+31-36#
#y=(x-6)^2-5#

Recall that the general equation of a quadratic equation written in vertex form is:

#y=a(x-h)^2+k#

where:
#h=#x-coordinate of the vertex
#k=#y-coordinate of the vertex

So in this case, the vertex is #(6,-5)#.