# How do you write f(x) = x^2 - 2x + 8 in vertex form?

Aug 15, 2016

In Vertex form $f \left(x\right) = {\left(x - 1\right)}^{2} + 7$.

#### Explanation:

$f \left(x\right) = {x}^{2} - 2 x + 8 = {\left(x - 1\right)}^{2} - 1 + 8 = {\left(x - 1\right)}^{2} + 7$.Vertex is at $\left(1 , 7\right)$ graph{x^2-2x+8 [-40, 40, -20, 20]}[Ans]

Aug 15, 2016

$f \left(x\right) = {\left(x - 1\right)}^{2} + 7 \text{ }$ Vertex is at $\left(1 , 7\right)$

#### Explanation:

Use the method of "completing the square."

${x}^{2} \textcolor{red}{- 2} x \textcolor{red}{+ 1} \textcolor{b l u e}{- 1} + 8 \text{ Add on half of (-2) squared}$

$\textcolor{red}{{\left(\frac{- 2}{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1}$

But we may not change the value of the expression by just adding 1.
1 must be subtracted as well.
$\textcolor{red}{+ 1} \textcolor{b l u e}{- 1} = 0$

${x}^{2} - 2 x + 1$ is a perfect square

$\left({x}^{2} - 2 x + 1\right) - 1 + 8 = {\left(x - 1\right)}^{2} + 7$

$f \left(x\right) = {\left(x - 1\right)}^{2} + 7 \text{ is the vertex form}$