How to complete the square to find vertex and x-intercept of parabola y=2x^2-11x+2

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The equation of a parabola in vertex form is `y = a(x – h)^2+ k` .

For the parabola `y = 2x^2 - 11x + 2` the vertex can be determined by expressing it in the form `y = a(x – h)^2+ k`

`y = 2x^2 - 11x + 2`

=> `2(x^2 - (11/2)x) + 2`

=> `2*(x^2 - 2*(11/4)*x + 121/16) + 2 - 121/8`

=> `2*(x - (11/4))^2 - 105/8`

The vertex of the parabola is `(11/4,- 105/8)`

The x-intercept is the solution of 2x^2 - 11x + 2 = 0

x1 = `11/4 + sqrt(121 - 16)/4`

=> `11/4 + sqrt 57/4`

x2 = `11/4 - sqrt 57/4`

The vertex of the parabola is `(11/4, -105/8)` and the x-intercepts are `11/4 + sqrt 57/4` and `11/4 - sqrt 57/4`

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