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

1 Answer
Pythagoras
Nov 29, 2015

y=-x^2+2x-3y=−x2+2x−3

Explanation:

The standard form is
y=ax^2 +bx +cy=ax2+bx+c
(where aa, bb and cc 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)^2y−4=−(x−1)2
which becomes:
y-4=-(x^2-2x+1)y−4=−(x2−2x+1)
(note that I still keep the minus sign in front of the parenthesis)

y-4=-x^2+2x-1y−4=−x2+2x−1

And pass the -4 on the "other side":
y=-x^2+2x-1+4y=−x2+2x−1+4

giving you:
y=-x^2+2x-3y=−x2+2x−3
which is in standard form with
a=-1a=−1, b=2b=2 and c=-3c=−3

Related questions
See all questions in Polynomials in Standard Form
Impact of this question
2551 views around the world
You can reuse this answer
Creative Commons License