which equation is the inverse of ((x - 4)^2 - \frac{2}{3} = 6y - 12)?\n\n(\bigcirc) (y = \frac{1}{6}x^2…

which equation is the inverse of ((x - 4)^2 - \frac{2}{3} = 6y - 12)?\n\n(\bigcirc) (y = \frac{1}{6}x^2 - \frac{4}{3}x + \frac{43}{9})\n\n(\bigcirc) (y = 4 pm sqrt{6x - \frac{34}{3}})\n\n(\bigcirc) (y = -4 pm sqrt{6x - \frac{34}{3}})\n\n(\bigcirc) (-(x - 4)^2 - \frac{2}{3} = -6y + 12)

which equation is the inverse of ((x - 4)^2 - \frac{2}{3} = 6y - 12)?\n\n(\bigcirc) (y = \frac{1}{6}x^2 - \frac{4}{3}x + \frac{43}{9})\n\n(\bigcirc) (y = 4 pm sqrt{6x - \frac{34}{3}})\n\n(\bigcirc) (y = -4 pm sqrt{6x - \frac{34}{3}})\n\n(\bigcirc) (-(x - 4)^2 - \frac{2}{3} = -6y + 12)

Answer

Explanation:

Step1: Swap x and y

To find the inverse, we first swap ( x ) and ( y ) in the original equation ((x - 4)^2-\frac{2}{3}=6y - 12). So we get ((y - 4)^2-\frac{2}{3}=6x - 12).

Step2: Solve for y

First, isolate the squared term: [ (y - 4)^2=6x - 12+\frac{2}{3} ] Simplify the right - hand side: [ 6x-12+\frac{2}{3}=6x-\frac{36}{3}+\frac{2}{3}=6x-\frac{34}{3} ] Then take the square root of both sides: [ y - 4=\pm\sqrt{6x-\frac{34}{3}} ] Finally, solve for ( y ): [ y = 4\pm\sqrt{6x-\frac{34}{3}} ]

Answer:

( y = 4\pm\sqrt{6x-\frac{34}{3}} ) (the second option: ( y = 4\pm\sqrt{6x-\frac{34}{3}} ))