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

##### 1 Answer

Dec 16, 2015

vertex

#### 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:

So in this case, the vertex is