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

2 Answers
Jim G.
Aug 6, 2018

#"vertex "=(0,-11)#

Explanation:

#"expand and rearrange into standard form"#

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

#y=x^2-2x+1+2x-12#

#y=x^2-11#

#"A quadratic in the form "y=ax^2+c#

#"has it's vertex at "(0,c)#

#"this has it's vertex at "(0,-11)#
graph{x^2-11 [-40, 40, -20, 20]}

Mark D.
Aug 6, 2018

#y=(x-1)^2+2x-12#

Expand the brackets

#y=x^2-2x+1+2x-12#

#y=x^2-11#

The parabola #y=x^2# is a #uu# 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 #(-b)/(2a)# when the equation is in the form #y=ax^2+bx+c#

From #y=x^2-11 a=1 and b=0#

#-0/1=0# put #x=0# into the equation, #y=-11#
(0,-11) is your vertex

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