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Anna55
05-25-2011, 12:08 PM
The parabola y = –x^2 + 4 has vertex P and intersects the x-axis at A and B.The parabola is translated from its original position so that its vertex moves along the line y = x + 4 to the point Q. In this position, the parabola intersects the x-axis at B and C. Determine the coordinates of C.

The parabola y = –x^2 + 4 has vertex P(0, 4) and intersects the x-axis at A(– 2, 0) and B(2, 0). The intercept B(2, 0) has its pre-image, B? on the parabola y = –x^2 + 4. To find B? , we find the point of intersection of the line passing through B(2, 0), with slope 1, and the parabola y = –x^2 + 4.

The equation of the line is y = x – 2.
Intersection points, x – 2 = –x^2 + 4
x^2+x-6=0
(x+3) (x-2)
Therefore, x = – 3 or x = 2.
For x = – 3, y = – 3 – 2 = – 5. Thus B? has coordinates (– 3, – 5).
If (– 3, – 5)?(2, 0) then the required general translation mpping y = –x^2 + 4 onto the
parabola with vertex Q is (x, y)?(x + 5, y + 5).

I do not understand the things in bold. Can you please explain?

galactus
05-27-2011, 04:53 PM
If we graph several of the shifted parabolas, it is easier to see. It is called a 'mapping'.

A picture is good for this to see what is going on. Here is a graph with two parabolas running down the line y=x+4.

See the intersection they mention, B'(-3,-5)?. It intersects the original parabola, and the translated one as the vertex moves along the line y=x+4.

The upper parabola has vertex (0,4)\rightarrow (5,9) per the translation given.

See a wee bit better now?.

Anna55
06-01-2011, 02:17 AM
Thank you galactus! The graphs you drew makes things much easier to understand!