1. ## 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?

2. 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?

3. ## I see..

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?

4. 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

5. ## Will this help?

You can draw it using "Function Drawing" (:
[I'd be more than happy to see if that worked out]

6. ## ok..........

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.

7. ## thanks........

Originally Posted by ShaiDuvdevani
You can draw it using "Function Drawing" (:
[I'd be more than happy to see if that worked out]