How do you write y=3(x-1)^2+5 in standard form?

Jul 25, 2017

See a solution process below:

Explanation:

First, we need to expand the squared term using this rule:

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

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

$y = 3 {\left(\textcolor{red}{x} - \textcolor{b l u e}{1}\right)}^{2} + 5$

$y = 3 \left({\textcolor{red}{x}}^{2} - \left[2 \cdot \textcolor{red}{x} \cdot \textcolor{b l u e}{1}\right] + {\textcolor{b l u e}{1}}^{2}\right) + 5$

$y = 3 \left({x}^{2} - 2 x + 1\right) + 5$

Next, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$y = \textcolor{red}{3} \left({x}^{2} - 2 x + 1\right) + 5$

$y = \left(\textcolor{red}{3} \times {x}^{2}\right) - \left(\textcolor{red}{3} \times 2 x\right) + \left(\textcolor{red}{3} \times 1\right) + 5$

$y = 3 {x}^{2} - 6 x + 3 + 5$

Now, combine like terms:

$y = 3 {x}^{2} - 6 x + \left(3 + 5\right)$

$y = 3 {x}^{2} - 6 x + 8$

Jul 25, 2017

$y = 3 {x}^{2} + 6 x + 8$

Explanation:

$y = 3 {\left(x - 1\right)}^{2} + 5$.
To convert the vertex form to standard form, we develop the vertex form:
$y = 3 \left({x}^{2} - 2 x + 1\right) + 5 = 3 {x}^{2} - 6 x + 3 + 5$
$y = 3 {x}^{2} - 6 x + 8$