Answer:
The answer is
[tex]y = \pm\sqrt{ \frac{1}{2}x - 1 } [/tex]
Step-by-step explanation:
[tex]y = {2x}^{2} + 2[/tex]
Interchange the terms
That's x becomes y and y becomes x
[tex]x = {2y}^{2} + 2[/tex]
Next solve for y
Send 2 to the other side of the equation
[tex] {2y}^{2} = x - 2[/tex]
Divide both sides by 2
[tex] \frac{ {2y}^{2} }{2} = \frac{x - 2}{2} \\ {y}^{2} = \frac{x}{2} - \frac{2}{2} \\ {y}^{2} = \frac{1}{2} x - 1[/tex]
Find the square root of both sides to make y stand alone
That's
[tex] \sqrt{ {y}^{2} } = \sqrt{ \frac{1}{2}x - 1 } [/tex]
We have the final answer as
[tex]y = \pm\sqrt{ \frac{1}{2}x - 1 } [/tex]
Hope this helps you
