Find a parabola with equation y=ax^2+bx+c that has a slope 10 at x=1, slope -26 at x=-1, and passes through the point (2,29)

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  y = 9x^2 -8x +9

Step-by-step explanation:

The given equation has derivative ...

  y' = 2ax +b

The requirements on slope give rise to two equations:

  2a(1) +b = 10

  2a(-1) +b = -26

Adding these equations together gives ...

  2b = -16   ⇒   b = -8

Then we have ...

  2a -8 = 10

  a = (10 +8)/2 = 9

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The given point lets us find the constant term c.

  y = 9x^2 -8x +c

  c = y -(9x -8)x = 29 -(9(2) -8)(2) = 29 -20 = 9

The equation of the parabola is ...

  y = 9x^2 -8x +9

to complete this square you have to rotate about the origin 270 degrees clockwise.