anonymous
  • anonymous
how do you rotate a function to a certain angle? say y = x^2+1 to 45degree
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you can draw a new XY coodinate system and center the graph about the line y=x|dw:1333574057443:dw|
anonymous
  • anonymous
I think you want the new equation also huh?
anonymous
  • anonymous
Well, that would require a bit more work...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
well yes, of course, so any help?
KingGeorge
  • KingGeorge
One of a couple problems that I'm seeing, is that \(x^2+1\) isn't a function anymore if you rotate it even the tiniest bit.
anonymous
  • anonymous
ok yes it would be broken into various funtions, so how are they?
KingGeorge
  • KingGeorge
I suppose one option would be to simply change the basis/variables. You can create new variables s, t such that the s axis is 45 degrees off the positive x-axis and the t axis is 45 degrees off the positive y-axis. Then write you equation as \(t=s^2+1\) and convert back to x, y after that.
KingGeorge
  • KingGeorge
You would still run into problems about the domain when converting though.
KingGeorge
  • KingGeorge
I think your change of variables would look something like \[s=\sqrt{2x^2}\]\[t=\sqrt{2y^2}\]
anonymous
  • anonymous
I don't think that will do
KingGeorge
  • KingGeorge
Not that I disagree with you, but why won't it work?
anonymous
  • anonymous
never mind, I have found an answer
KingGeorge
  • KingGeorge
Out of curiosity, what was it?
anonymous
  • anonymous
first there is: x^2 - 2xy + y^2 - x * sqrt(2) - y * sqrt(2) = 0 and when solved for y there is: y = [(sqrt(2) + 2x) +/- sqrt(4x^2 + 2 + 4 sqrt(2) x - 4x^2 + 4 sqrt(2) x)] / 2 or y = [sqrt(2) + 2x +/- sqrt(8 sqrt (2) x + 2)]/2.

Looking for something else?

Not the answer you are looking for? Search for more explanations.