Question

Equation Parabola `x^2-4x+2=2y`?

Answer 1

Given: `x^2-4x+2=2y`

Divide both sides of the equation by 2:

`y = 1/2x^2-2x+1`

The above equation is now in the standard form `y = ax^2+bx+c` where `a = 1/2`, `b=-2`, and `c = 1`; its roots can be found using the quadratic formula:

`x = (-b+-sqrt(b^2-4(a)(c)))/(2a)`

Substituting the values for, a, b, and c:

`x = (-(-2)+-sqrt((-2)^2-4(1/2)(1)))/(2(1/2))`

`x = (2+-sqrt(4-2))/1`

`x = 2+-sqrt2`

`x = 2+sqrt2` and `x = 2-sqrt2`