# How do you write f(x) + 9 = 2(x - 1)^2 in standard form?

Jun 16, 2017

See a solution process below:

#### Explanation:

First, we need to expand the terms in parenthesis. This term is a special form of the quadratic.:

(a - b)^2 = a^2 - 2ab + b^2

Substituting $x$ for $a$ and $1$ for $b$ gives:

$f \left(x\right) + 9 = 2 {\left(x - 1\right)}^{2}$

$f \left(x\right) + 9 = 2 \left({x}^{2} - \left(2 \cdot x \cdot 1\right) + {1}^{2}\right)$

$f \left(x\right) + 9 = 2 \left({x}^{2} - 2 x + 1\right)$

Next, we eliminate the parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$f \left(x\right) + 9 = \textcolor{red}{2} \left({x}^{2} - 2 x + 1\right)$

$f \left(x\right) + 9 = \left(\textcolor{red}{2} \cdot {x}^{2}\right) - \left(\textcolor{red}{2} \cdot 2 x\right) + \left(\textcolor{red}{2} \cdot 1\right)$

$f \left(x\right) + 9 = 2 {x}^{2} - 4 x + 2$

Now, subtract $\textcolor{red}{9}$ from each side of the function to complete the transformation to standard form:

$f \left(x\right) + 9 - \textcolor{red}{9} = 2 {x}^{2} - 4 x + 2 - \textcolor{red}{9}$

$f \left(x\right) + 0 = 2 {x}^{2} - 4 x - 7$

$f \left(x\right) = 2 {x}^{2} - 4 x - 7$

Jun 16, 2017

$f \left(x\right) = 2 {x}^{2} - 4 x - 7$

#### Explanation:

$\text{the standard form of a parabola is}$

color(red)(bar(ul(|color(white)(2/2)color(black)(f(x)=ax^2+bx+c;a!=0)color(white)(2/2)|)))

$\text{expand the bracket using FOIL}$

$\Rightarrow f \left(x\right) + 9 = 2 \left(x - 1\right) \left(x - 1\right)$

$\textcolor{w h i t e}{\Rightarrow f \left(x\right) + 9} = 2 \left({x}^{2} - 2 x + 1\right)$

$\textcolor{w h i t e}{\Rightarrow f \left(x\right) + 9} = 2 {x}^{2} - 4 x + 2$

$\text{subtract 9 from both sides}$

$\Rightarrow f \left(x\right) = 2 {x}^{2} - 4 x + 2 - 9$

$\textcolor{w h i t e}{\Rightarrow f \left(x\right)} = 2 {x}^{2} - 4 x - 7 \leftarrow \textcolor{red}{\text{ in standard form}}$