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:

$(5;3)$

$(7;7)$

$(13;11)$

$(37;19)$

How do you graph 8(x-5)=(y-3)^28(x-5)=(y-3)^2?