How do you write y = x^2 + 6x + 10 in vertex form?

Aug 14, 2016

$\textcolor{g r e e n}{y = {\left(x + 3\right)}^{2} + 1}$

Explanation:

$y = {x}^{2} + 6 x + 10$

$y = a {x}^{2} + b x + c \text{ This is the standard form}$

$\text{The vertex form is expressed as }$

$y = a {\left(x - h\right)}^{2} - k$

$F = \left(h , k\right)$

$F = \left(h , k\right) \text{represents the focus coordinates}$

$\text{Solution -1:}$

$y = {x}^{2} + 6 x + 9 + 1$

$\textcolor{g r e e n}{y = {\left(x + 3\right)}^{2} + 1}$

$h = - 3$

$k = 1$

$\text{Solution-2:}$

$h = - \frac{b}{2 a} = - \frac{6}{2 \cdot 1} = - 3$

$k = {\left(- 3\right)}^{2} + 6 \cdot \left(- 3\right) + 10 = 9 - 18 + 10 = 1$

$a = 1$

$\text{The vertex form:}$

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

$\textcolor{g r e e n}{y = {\left(x + 3\right)}^{2} + 1}$