Find the parabola with equation y = ax^2 + bx whose tangent line at (3, 6) has equation y = 8x − 18.

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The parabola with equation y = ax^2 + bx has a tangent at (3, 6) given by y = 8x - 18

As the point (3, 6) lies on the parabola

6 = 9a + 3b ...(1)

The derivative of y = ax^2 + bx is y' = 2ax + b. The value of this at any point is the slope of the tangent at that point. The slope of y = 8x - 18 is 8

=> 6a + b = 8 ...(2)

(1) - 3*(2)

=> 9a + 3b - 18a - 3b = 6 - 24

=> -9a = -18

=> a = 2

12 + b = 8

=> b = -4

**The equation of the parabola is y = 2x^2 - 4x**

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