How do you graph #8(x-5)=(y-3)^2#?

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Answer 1 · 2015-08-28T21:44:22

graph{sqrt(8(x-5))+3 [-1.5, 44, -2.5, 21]}

To graph #8(x-5)=(y-3)^2# you will need to isolate #y# to find its value with respect to #x#:

#8(x-5)=(y-3)^2 rarr sqrt(8(x-5))=y-3#

#rarr y=sqrt(8(x-5))+3#

From here you can conclude that #y#'s domain will be #x=[5;+oo[# and its range will be #y=[3;+oo[#

Here are some points through which the curve will go:

#(5;3)#

#(7;7)#

#(13;11)#

#(37;19)#