How do you graph $y \geq 2 x^{2} - 2 x - 5$ $y>=2x^2-2x-5$ ?

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Answer 1 · 2017-01-01T02:19:09

The graph is inserted.

graph{y>=2x^2-2x-5 [-6, 6, -12, 12]}

$y = 2 x^{2} - 2 x - 5 = 2 {(x - \frac{1}{2})}^{2} - \frac{11}{2}$ $y = 2x^2-2x-5=2(x-1/2)^2-11/2$ , giving the standard form

${(x - \frac{1}{2})}^{2} = 4 (\frac{1}{8}) (y + \frac{11}{2})$ $(x-1/2)^2=4(1/8)(y+11/2)$

The vertex ( least-y point ) of this parabola is $(\frac{1}{2}, - \frac{11}{2})$ $(1/2, -11/2)$ .

For any point (x, y) on or above this parabola,

y >= 2x^2-2x-5>=-11/2#.

How do you graph y>=2x^2-2x-5y≥2x2−2x−5y>=2x^2-2x-5?