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

1 Answer
Nov 29, 2015

Answer:

#y=-x^2+2x-3#

Explanation:

The standard form is
#y=ax^2 +bx +c#
(where #a#, #b# and #c# are some numbers)
So in this case, you just need to open up the parentheses of the right side of the equation and then rearrange the terms.

We have:
#y-4 = -(x-1)^2#
which becomes:
#y-4=-(x^2-2x+1)#
(note that I still keep the minus sign in front of the parenthesis)

#y-4=-x^2+2x-1#

And pass the -4 on the "other side":
#y=-x^2+2x-1+4#

giving you:
#y=-x^2+2x-3#
which is in standard form with
#a=-1#, #b=2# and #c=-3#