What is the the vertex of y = (x-1)^2+2x-12?

Aug 6, 2018

$\text{vertex } = \left(0 , - 11\right)$

Explanation:

$\text{expand and rearrange into standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$y = {x}^{2} - 2 x + 1 + 2 x - 12$

$y = {x}^{2} - 11$

$\text{A quadratic in the form } y = a {x}^{2} + c$

$\text{has it's vertex at } \left(0 , c\right)$

$\text{this has it's vertex at } \left(0 , - 11\right)$
graph{x^2-11 [-40, 40, -20, 20]}

Aug 6, 2018

$y = {\left(x - 1\right)}^{2} + 2 x - 12$

Expand the brackets

$y = {x}^{2} - 2 x + 1 + 2 x - 12$

$y = {x}^{2} - 11$

The parabola $y = {x}^{2}$ is a $\cup$ curve with the vertex (a minimum) at the origin (0,0)
$y = {x}^{2} - 11$ is this same curve but translated 11 units down the y axis so the vertex (again a minimum) is at (0,-11)

Another method:
To find the x coordinate of the vertex use $\frac{- b}{2 a}$ when the equation is in the form $y = a {x}^{2} + b x + c$

From $y = {x}^{2} - 11 a = 1 \mathmr{and} b = 0$

$- \frac{0}{1} = 0$ put $x = 0$ into the equation, $y = - 11$