| 
11-10-2008, 10:35 PM
| | MHF Contributor | | Join Date: Jul 2008 Location: Manhattan
Posts: 1,232
Country:  Thanks: 870
Thanked 16 Times in 12 Posts
| |  Parabola
Draw y = (x^2/2) - 3x + 4 on the interval [0, 6].
What are the coordinates of the turning point of the image of the graph of the given parabola after a reflection in the line y = 4?
I graphed the function and found the point to be (3, 4).
Is this right? If not, can you explain why? |
11-10-2008, 11:05 PM
| | Member | | Join Date: Oct 2008 Location: Seattle, Washington
Posts: 63
Country: Thanks: 2
Thanked 19 Times in 19 Posts
| |
No, that is not correct.
Since the coordinates of the vertex point are (3, -1/2) before reflecting, this point is located 4.5 units below the line y = 4.
Therefore, after reflecting, the vertex point must be located 4.5 units above the line y = 4.
Did you draw a picture of the reflected graph?
Last edited by mmm4444bot; 11-10-2008 at 11:58 PM.
| | The following users thank mmm4444bot for this useful post: | |  |
11-11-2008, 11:07 PM
| | MHF Contributor | | Join Date: Jul 2008 Location: Manhattan
Posts: 1,232
Country: Thanks: 870
Thanked 16 Times in 12 Posts
| | I see..
Quote:
Originally Posted by mmm4444bot  No, that is not correct.
Since the coordinates of the vertex point are (3, -1/2) before reflecting, this point is located 4.5 units below the line y = 4.
Therefore, after reflecting, the vertex point must be located 4.5 units above the line y = 4.
Did you draw a picture of the reflected graph? | The point would be (3, 17/2), right? |
11-12-2008, 12:56 AM
| | Member | | Join Date: Oct 2008 Location: Seattle, Washington
Posts: 63
Country: Thanks: 2
Thanked 19 Times in 19 Posts
| |
Yes, that is correct. Any time you reflect an image across a line, all of the points on the reflected image must be the same distance from the axis of reflection as their corresponding points on the original image are. ~ Mark  | | The following users thank mmm4444bot for this useful post: | | |
11-12-2008, 07:09 AM
| | Newbie | | Join Date: Nov 2008
Posts: 4
Country:  Thanks: 0
Thanked 2 Times in 2 Posts
| | Will this help?
You can draw it using "Function Drawing" (:
[I'd be more than happy to see if that worked out]
link: Drawing Tools | | The following users thank ShaiDuvdevani for this useful post: | | |
11-12-2008, 06:38 PM
| | MHF Contributor | | Join Date: Jul 2008 Location: Manhattan
Posts: 1,232
Country: Thanks: 870
Thanked 16 Times in 12 Posts
| | ok..........
Quote:
Originally Posted by mmm4444bot Yes, that is correct. Any time you reflect an image across a line, all of the points on the reflected image must be the same distance from the axis of reflection as their corresponding points on the original image are. ~ Mark | Thank you very much. |
11-12-2008, 06:38 PM
| | MHF Contributor | | Join Date: Jul 2008 Location: Manhattan
Posts: 1,232
Country: Thanks: 870
Thanked 16 Times in 12 Posts
| | thanks........
Quote:
Originally Posted by ShaiDuvdevani You can draw it using "Function Drawing" (:
[I'd be more than happy to see if that worked out]
link: Drawing Tools | Thanks for your input. |
11-12-2008, 10:52 PM
| | Newbie | | Join Date: Nov 2008
Posts: 4
Country: Thanks: 0
Thanked 2 Times in 2 Posts
| | Thanks
Youre most welcome (:
Hope that worked out for you | | The following users thank ShaiDuvdevani for this useful post: | | | | Thread Tools | | | | Display Modes |  Linear Mode | 
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts
HTML code is Off | | | All times are GMT -7. The time now is 06:19 PM. | | |
 | |  |
