How do I put y=(x-2)^2-4 in the form y=ax^2+bx+c
Thanks for any help.
2007-11-14 04:39:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y=(x-2)²2-4 in the form y=ax²2+bx+c ?
I don't get where the '-4x' is coming from in the answers im getting.
2007-11-14 05:21:21 · update #1
y = (x - 2)(x - 2) - 4
y = x² - 4x + 4 - 4
y = x² - 4x
a = 1
b = - 4
c = 0
2007-11-14 07:38:33 · answer #1 · answered by Como 7 · 2⤊ 1⤋
First, expand the equation.
y=(x-2)^2-4
y=x^2 - 4x + 4 - 4
y =x^2 - 4x + 0
The above equation is in the form,
y=ax^2+bx+c
where, a = 1 , b= -4 and c= 0
2007-11-14 04:44:31 · answer #2 · answered by crashbird 2 · 0⤊ 1⤋
You need to carry out the square operation.
y=(x-2)^2-4
=x²-4x+4-4
2007-11-14 04:44:09 · answer #3 · answered by DWRead 7 · 0⤊ 2⤋
y=x^2-4x+2-4
y=x^2-4x-2
2007-11-14 04:43:55 · answer #4 · answered by Anonymous · 0⤊ 1⤋
y=(x-2)^2-4
just expand
y=x^2-4x+4-4
y=x^2-4x+0
y=x^2-4x
2007-11-14 04:54:23 · answer #5 · answered by Siva 5 · 0⤊ 1⤋