How do you find the vertex and intercepts for $x=1+y-y^2$ ?

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Answer 1 · 2016-06-23T11:28:48

vertex at $x = 5/4, y = 1/2$

intercepts $x = 0, y = (1 pm sqrt(5))/(2)$

$x=1+y-y^2$
$x=-(y^2 -y -1)$
$x=-( (y- 1/2 )^2 - 1/4 -1)$
$x=-( (y- 1/2 )^2 - 5/4)$
$x= -(y- 1/2 )^2 + 5/4$

vertex at $x = 5/4, y = 1/2$

$x = 0 \implies 1+y-y^2 = 0$

$y^2 - y - 1 = 0$

$y = (-(-1) pm sqrt((-1)^2 - 4(1)(-1) ))/(2*1)$

$= (1 pm sqrt(5))/(2)$ = intercepts

How do you find the vertex and intercepts for x=1+y-y^2x=1+y-y^2?