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

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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 $8 (x - 5) = {(y - 3)}^{2}$ $8(x-5)=(y-3)^2$ you will need to isolate $y$ $y$ to find its value with respect to $x$ $x$ :

$8 (x - 5) = {(y - 3)}^{2} \to \sqrt{8 (x - 5)} = y - 3$ $8(x-5)=(y-3)^2 rarr sqrt(8(x-5))=y-3$

$\to y = \sqrt{8 (x - 5)} + 3$ $rarr y=sqrt(8(x-5))+3$

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

Here are some points through which the curve will go:

$(5; 3)$ $(5;3)$

$(7; 7)$ $(7;7)$

$(13; 11)$ $(13;11)$

$(37; 19)$ $(37;19)$

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