1. ## in "layman's terms"....

I am beginning to work with Polynomial, Rational, and Quadratic functions...

I can understand most of it but I am having issues trying to understand who it works if written like this,
$f(x)=x^2+2x$

I am told that the first thing to do is completing the Sqr, by adding one and subtracting one, thus;

$f(x)=x^2+2x+1-1$

$x^2+2x+1=(x+1)^2-1$

$x+1=0$

$x=-1$

and so my vertex is (-1,-1)

but now how about the $X$ and the $Y$ coordinates?

Thank you and sorry for such a long post.

2. Originally Posted by Morzilla
I am beginning to work with Polynomial, Rational, and Quadratic functions...

I can understand most of it but I am having issues trying to understand who it works if written like this,
$f(x)=x^2+2x$

I am told that the first thing to do is completing the Sqr, by adding one and subtracting one, thus;

$f(x)=x^2+2x+1-1$

$x^2+2x+1=(x+1)^2-1$

$x+1=0$

$x=-1$

and so my vertex is (-1,-1)

but now how about the $X$ and the $Y$ coordinates?

Thank you and sorry for such a long post.

x and y coordinates for what?

you didn't have to solve x + 1 = 0 explicitly.

if a quadratic is expressed as $y = a(x - h)^2 + k$, then the vertex is given by $(h,k)$

3. Originally Posted by Jhevon
x and y coordinates for what?

you didn't have to solve x + 1 = 0 explicitly.

if a quadratic is expressed as $y = a(x - h)^2 + k$, then the vertex is given by $(h,k)$
soooo........how does $x^2$ and $2x$ fit into $y = a(x - h)^2 + k$

Thanks

4. Originally Posted by Morzilla
soooo........how does $x^2$ and $2x$ fit into $y = a(x - h)^2 + k$

Thanks
i'm not sure i got your question. we completed the square for f(x) = x^2 + 2x to get it into the above form. in that form, a = 1, h = -1 and k = -1