# How do you find the vertex of y= x^2-8x+5?

Aug 1, 2015

The simplest method for finding the vertex of the given polynomial is to convert it into "vertex form" to find the vertex at $\left(4 , - 11\right)$

#### Explanation:

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

If we can convert this into vertex form:
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$for some constants $a$ and $b$
then the vertex will be at $\left(a , b\right)$

The easiest way to do this conversion is via a "completion of the square method"

If ${x}^{2} - 8 x$ are the first 2 terms of a squared binomial
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$note: ${\left(m + n\right)}^{2} = \left({m}^{2} + 2 m n + {n}^{2}\right)$
then the third term needs to be ${\left(\frac{- 8 x}{2 x}\right)}^{2} = 16$

So we add (and then subtract $16$ to "complete the square"

$y = \left({x}^{2} - 8 x + 16\right) + 5 - 16$

Simplifying:
$y = {\left(x - 4\right)}^{2} + \left(- 11\right)$
$\textcolor{w h i t e}{\text{XXXX}}$which is our desired "vertex form" (with $m = 1$)

So the vertex is at $\left(4 , - 11\right)$

Aug 1, 2015

Find vertex of f(x) = x^2 - 8x + 5

Ans: Vertex (4, -11)

#### Explanation:

x-coordinate of vertex: $x = - \frac{b}{2 a} = \frac{8}{2} = 4$

y-coordinate of vertex: $y = f \left(4\right) = {4}^{2} - 8 \left(4\right) + 5 = - 11$