# How do you write y=-2(x+1)^2+3?

Jun 22, 2018

See a solution process below:

#### Explanation:

First, use this rule for quadratics to expand the term within the parenthesis:

${\left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right)}^{2} = \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right) \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right) = {\textcolor{red}{a}}^{2} + 2 \textcolor{red}{a} \textcolor{b l u e}{b} + {\textcolor{b l u e}{b}}^{2}$

$y = - 2 {\left(\textcolor{red}{x} + \textcolor{b l u e}{1}\right)}^{2} + 1$

$y = - 2 \left(\textcolor{red}{x} + \textcolor{b l u e}{1}\right) \left(\textcolor{red}{x} + \textcolor{b l u e}{1}\right) + 1$

$y = - 2 \left({\textcolor{red}{x}}^{2} + \left(2 \cdot \textcolor{red}{x} \cdot 1\right) + {\textcolor{b l u e}{1}}^{2}\right) + 1$

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

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

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

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

y = -2x^2 + (-4x) + (-2)) + 1

$y = - 2 {x}^{2} - 4 x - 2 + 1$

Now, we can combine the common terms, for this problem the constants:

$y = - 2 {x}^{2} - 4 x - 1$