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