Find the inverse function of a half parabola

Question

<sub>Help me find the inverse function of a half parabola

-2x^2 + x - 2, where x > 1/4
the homework system says f^(-1) (x) = -1/4 + √(-(x + 15/8) / 2) is not correct</sub>

Answer 1

y = -2x²+x-2

 

y = -2(x² - (1/2)x) - 2

 

y = -2(x² - (1/2)x + 1/16 - 1/16) - 2

 

y = -2(x-1/4)² -15/8   (this is a parabola opening downward with vertex                                     (1/4, -15/8))

 

y = -2(x-1/4)²-15/8, x≥1/4 is the right half of the parabola

 

Find the inverse:  solve for x:  y + 15/8 = -2(x-1/4)²

 

                                           (x-1/4)² = (y+15/8)/(-2)

 

                                            x-1/4 = ±√[(y+15/8)/(-2)]

 

                                                 x = 1/4 ± √[(y+15/8)/(-2)]

 

Since x≥1/4, x = 1/4 + √[(y+15/8)/(-2)]

 

Switch x and y to get y = f^-1(x) = 1/4 + √[(x+15/8)/(-2)], x≤-15/8

 

Answer 2

It should be +1/4, not minus 1/4.

 

We complete the square and use (sqrt{2}x - sqrt{2}/4)^2 = 2x - x + 1/8

So negate the left side and subtract 15/8:

 

-y-15/8 = (sqrt{2}x - sqrt{2}/4)^2

 

We want the positive square root since we are considering the part with x>1/4, so

 

sqrt{-8y - 15}/2sqrt{2} = sqrt{2}x - sqrt{2}/4

sqrt{-8y - 15}/2sqrt{2} + sqrt{2}/4 = sqrt{2}x         *** where the error was likely made

sqrt{-8y - 15}/4 + 1/4 = x

[sqrt{-8x - 15} + 1]/4 = y