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

2 Answers
Mar 9, 2016

#(5/2,7/4)#

Explanation:

First expand the equation to get it into standard form, then convert into vertex form by completing the square.

#y = (x^2 - 4x - 2x +8) +x#
#y = x^2-5x +8#

#y = (x-5/2)^2 -25/4 +8#

#y = (x-5/2)^2 +7/4#

The vertex is #(5/2,7/4)# which is the point where the bracketed term is zero and therefore the expression is at its minimum.

Mar 9, 2016

A related but very slightly different approach

#color(green)("Vertex " ->" "(x,y)" "->" "(5/2,7/4)#

Explanation:

An alternative approach. It does in fact incorporate part of the process of constructing the vertex equation.

Multiply out the brackets

#y=x^2-6x+8+x#

#y=x^2-5x+8#

Consider the #-5# from #-5x#

Apply# (-1/2)xx(-5) = +5/2#

#color(blue)(x_"vertex"=5/2)#

By substitution

#color(blue)(y_("vertex")=(5/2)^2-5(5/2)+8= +7/4)#

#color(green)("Vertex " ->" "(x,y)" "->" "(5/2,+7/4)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("A word of caution")#

given that the standard form is# y=ax^2+bx+c#

When applying this approach you must have
#" "y=a(x^2+b/ax)+c#

So in fact#" "y_("vertex") = (-1/2)xx(b/a)#

In your question #a=1# so for that question

#" "color(brown)(y_("vertex") = (-1/2)xx(b/a))color(green)( -> (-1/2)xx(-5/1))#