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

Mar 9, 2016

$\left(\frac{5}{2} , \frac{7}{4}\right)$

#### Explanation:

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

$y = \left({x}^{2} - 4 x - 2 x + 8\right) + x$
$y = {x}^{2} - 5 x + 8$

$y = {\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4} + 8$

$y = {\left(x - \frac{5}{2}\right)}^{2} + \frac{7}{4}$

The vertex is $\left(\frac{5}{2} , \frac{7}{4}\right)$ 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

$y = {x}^{2} - 6 x + 8 + x$

$y = {x}^{2} - 5 x + 8$

Consider the $- 5$ from $- 5 x$

Apply$\left(- \frac{1}{2}\right) \times \left(- 5\right) = + \frac{5}{2}$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \frac{5}{2}}$

By substitution

$\textcolor{b l u e}{{y}_{\text{vertex}} = {\left(\frac{5}{2}\right)}^{2} - 5 \left(\frac{5}{2}\right) + 8 = + \frac{7}{4}}$

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{red}{\text{A word of caution}}$

given that the standard form is$y = a {x}^{2} + b x + c$

When applying this approach you must have
$\text{ } y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

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))