# Finding stretch or compression

• Nov 29th 2010, 05:39 PM
Jubbly
Finding stretch or compression
Hey everyone I'm trying to figure out how to find a stretch or compression of a parabola. I have a parabola with a vertex of (1,3) and points are (0,5) and (2,5).
The equation I have so far is y = _ (x-1)^2+3, but i can't find the stretch/compression. Thanks for your help.
• Nov 29th 2010, 05:44 PM
Ackbeet
If you mean that the parabola you've found so far is

\$\displaystyle y=(x-1)^{2}+3,\$ then I would not agree that all three of

\$\displaystyle (1,3), (2,5), (0,5)\$ are on the parabola.

That is, I don't think you've found the correct formula yet.
• Nov 29th 2010, 06:01 PM
Jubbly
The parabola is a stretched parabola with the points (0,5) and (2,5). and a vertex of (1,3). I'm trying to find the stretc/compression of this parabola.
• Nov 30th 2010, 02:44 AM
Ackbeet
If you're trying to find the stretch or compression of the parabola, you must be comparing this parabola to a different parabola. You can fit a parabola to the three points you have there. And that would just be a simple parabola. To know if that parabola is stretched or compressed from another parabola, you would have to tell me what the equation for the other parabola is. Do you follow me? I guess what I'm saying is that the very word "stretch" implies that you're starting with one thing, and then stretching it into another thing. Thus, there are two things involved. I only see one thing at the moment.
• Nov 30th 2010, 03:32 AM
HallsofIvy
Ackbeet, there is a "_" in front of the square representing, I think, the unknown coefficient. I believe that Jubbly means \$\displaystyle y= a(x- 1)^2+ 3\$. The "-1" and "+ 3", of course, guarantee that the parabola passes through (1, 3). Putting x= 2 and y= 5 gives \$\displaystyle 5= a(2- 1)^2+ 3= a+ 3\$. Putting x= 0 and y= 5 gives \$\displaystyle 5= a(0- 1)^2+ 3= a+ 3\$ also