How do you complete the square to find the vertex for $y + 6 = x^{2} - 5 x$ $y + 6 = x^2 - 5x$ ?

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Answer 1 · 2015-06-07T23:07:56

$y + 6 = x^{2} - 5 x$ $y+6 = x^2-5x$

$y = x^{2} - 5 x - 6$ $y = x^2-5x-6$

$y = x^{2} - 5 x + \frac{25}{4} - 6 - \frac{25}{4}$ $y = x^2 -5x + color(blue) (25/4) -6 -color(blue)(25/4)$

$y = {(x - \frac{5}{2})}^{2} - \frac{24 + 25}{4}$ $y = (x-5/2)^2-(24+25)/4$

$y = {(x - \frac{5}{2})}^{2} - \frac{49}{4}$ $y = (x-5/2)^2 -49/4$

Vertex: $(\frac{5}{2}, - \frac{49}{4})$ $(5/2, -49/4)$

How do you complete the square to find the vertex for y+6=x2−5xy + 6 = x^2 - 5x?