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 BTony 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))