What is the vertex of # y= (x+1)^2-2x-4#?

1 Answer
Apr 29, 2016

Vertex form#" "y=(x+0)^2-3#

So the vertex is at #(x,y)->(0,-3)#

This is the same as #y=x^2-3#

Explanation:

There is an inherent #bx# term within #(x+1)^2#. Normally you would expect all of #bx# terms to be within the brackets. One is not! Consequently the brackets have to be expanded so that the excluded term of #-2x# can be incorporated with the term (hidden) in the brackets.

Expanding the brackets # y= (x^2+2x+1) -2x-4#

Combining terms:#" "y=x^2+0x-3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertex form")#

Standard form:#" "y=ax^2+bx+c" "# in your case #a=1#

Vertex form:#" "y=a(x+b/(2a))^2+c -a(b/(2a))^2#

But #b/(2a)=0" "# so #" "-a(b/(2a))^2=0#

#y=(x+0)^2-3" " ->" " y=x^2-3#