Hello

Question

Hello

igreen · Answer

What was your final answer?

anonymous · Answer

y=21. But I have the feeling I did the steps wrong.

igreen · Answer

No..that's incorrect..the equation you are given is a parabola..y = 21 is a completely straight horziontal line.

igreen · Answer

*horizontal

anonymous · Answer

So, how should I start the equation off?

igreen · Answer

Okay, first we're going to add 3 to both sides.

$y = x^2 -2x - 3$

$y + 3 = x^2 - 2x$

igreen · Answer

Now we complete the square.

igreen · Answer

$x^2 - 2x$

We take the '-2' and take half of it, which is -1, now we square -1.

$-1^2 \rightarrow -1 \times -1 \rightarrow 1$

So therefore we have:

$x^2 - 2x + 1$

Plug it back in:

$y + 3 + 2(1) = x^2 - 2x + 1$

Simplify:

$y + 3 + 2 = x^2 - 2x + 1$

$y + 5 = x^2 - 2x + 1$

anonymous · Answer

I got $$y=x^2(x+1)^2-4$$ as the answer.

igreen · Answer

No..that's not right..

igreen · Answer

$y + 5 = x^2 - 2x + 1$

Now we factor $x^2 - 2x + 1$.

$y + 5 = (x - 1)^2$

Now our last step is to subtract 5 to both sides, what do you get?

igreen · Answer

Sorry that's:

$y + 4 = (x - 1)^2$

Now subtract 4 to both sides..

anonymous · Answer

I got y=(x-1)^2-4

I subtracted 4 to both sides.

igreen · Answer

Yep, you got it.

igreen · Answer

Yes.

igreen · Answer

We completed the square.

igreen · Answer

http://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html

anonymous · Answer

Ah I see now! Thank you. 

You don't mind helping me out on finding the zero on this equation still?

igreen · Answer

You could just graph it to find the zero.

igreen · Answer

There's actually 2 of them. https://www.desmos.com/calculator/srjhfxxnjg