What is the the vertex of $x = {(y + 2)}^{2} - 5 y - 2$ $x = (y +2)^2 - 5y-2$ ?

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Answer 1 · 2018-06-14T14:04:45

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Answer 2 · 2018-06-14T14:10:02

$(\frac{7}{4}, \frac{1}{2})$ $(7/4 , 1/2)$

$x = {(y + 2)}^{2} - 5 y - 2$ $x = (y+2)^2-5y-2$

$x = (y^{2} + 4 y + 4) - 5 y - 2$ $x = (y^2 color(blue)(+4y) +4)color(blue)(-5y)-2$

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

$x = {(y - \frac{1}{2})}^{2} + \frac{7}{4}$ $x = (y- 1/2)^2 + 7/4$

$x - \frac{7}{4} = {(y - \frac{1}{2})}^{2}$ $x- 7/4 = (y - 1/2)^2$

$The vertex is at (\frac{7}{4}, \frac{1}{2})$ $"The vertex is at " (7/4,1/2)$

What is the the vertex of x=(y+2)2−5y−2x = (y +2)^2 - 5y-2 ?