2 posts
• Page **1** of **1**

. . .

. . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

To start with I graphed the equation 1+(x-1)^2. I see that the origin of the parabola is shifted one unit up and one unit to the right and the parabola faces upward. So method one applies the formula on a line in point slope form. So an arbitrary point a is applied and x and y are set to zero. The resulting equation is solved for a. I think I understand this far.

What I am still figuring is y=2( (sqrt 2) - 1 )x and y=-2( (sqrt2) + 1 )x

I don't see yet how those were derived fully.

Thanks in advance...

- sepoto
**Posts:**8**Joined:**Thu Sep 19, 2013 3:38 am

sepoto wrote:

. . . . .

plug sqrt[2] in for a in 1st line eqn:

y=2(sqrt[2]-1)(x-sqrt[2])+1+(sqrt[2]-1)^2

=2(sqrt[2]x-x-2+sqrt[2])+1+(2-2sqrt[2]+1)

=2sqrt[2]x-2x-4+2sqrt[2]+1+2-2sqrt[2]+1

=2sqrt[2]x-2x-4+1+2+1+2sqrt[2]-2sqrt[2]

=2sqrt[2]x-2x

=2(sqrt[2]x-x)

=2(sqrt[2]-1)x

the other one works the same way

- buddy
**Posts:**113**Joined:**Sun Feb 22, 2009 10:05 pm

2 posts
• Page **1** of **1**