Equation Parabola $x^{2} - 4 x + 2 = 2 y$ $x^2-4x+2=2y$ ?

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Equation Parabola $x^{2} - 4 x + 2 = 2 y$ $x^2-4x+2=2y$ ?

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Answer 1 · 2018-06-26T17:03:46

Given: $x^{2} - 4 x + 2 = 2 y$ $x^2-4x+2=2y$

Divide both sides of the equation by 2:

$y = \frac{1}{2} x^{2} - 2 x + 1$ $y = 1/2x^2-2x+1$

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 (a) (c)}}{2 a}$ $x = (-b+-sqrt(b^2-4(a)(c)))/(2a)$

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

$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (\frac{1}{2}) (1)}}{2 (\frac{1}{2})}$ $x = (-(-2)+-sqrt((-2)^2-4(1/2)(1)))/(2(1/2))$

$x = \frac{2 \pm \sqrt{4 - 2}}{1}$ $x = (2+-sqrt(4-2))/1$

$x = 2 \pm \sqrt{2}$ $x = 2+-sqrt2$

$x = 2 + \sqrt{2}$ $x = 2+sqrt2$ and $x = 2 - \sqrt{2}$ $x = 2-sqrt2$

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