Question

How do you complete the square with an equation like y = x^2 - 2x - 3? Please just show steps.

Answer 1

All right, we can complete the square as such:
\[
y=x^2-2x-3\\
\left(\frac{-2}{2}\right)^2=(-1)^2=1\implies\\
y=(x^2-2x+1)-3+1
\](We add unity outside since we just added unity inside) Factoring and simplifying gives us:
\[
y=(x-2x+1)-2\implies\\
y=(x-1)^2-2
\]Et voilá.

Answer 2

Dude thanks much. I was confused on the -2x because I thought you changed that. Thanks though.

Answer 3

Sure thing.

Answer 4

Wait, my bad, I messed up on one thing, it is supposed to subtract one outside:
\[
y=(x-1)^2-4
\]Not \(-2\).

Answer 5

Yea I was just about to say. I'm trying to graph it right now and just noticed the vertex and the y-intercept doesn't match up.

Answer 6

Yeah, here:
\[
y=x^2-2x-3\\
\left(\frac{-2}{2}\right)^2=(-1)^2=1\implies\\
y=(x^2-2x+1)-3-1\implies\\
y=(x-1)^2-4
\]

Answer 7

Couly you show me how you would do \[-3x ^{2} + 3x - 2\] please?

Answer 8

Sure:
\[
y=-3x^2+3x-2\implies\\
y=-3(x^2-x)-2
\]We know:
\[
\left(-\frac{1}{2}\right)^2=\frac{1}{4}
\]So:
\[
y=-3\left(x^2-x+\frac{1}{4}\right)-2+3\cdot\frac{1}{4}=-3\left(x^2-x+\frac{1}{4}\right)-\frac{5}{4}\implies\\
y=-3\left(x-\frac{1}{2}\right)^2-\frac{5}{4}
\]And, we're done.

Answer 9

Thanks. It is starting to make since now.

Answer 10

Cool, glad it is!