Question 814829
Not sure if it is called a critical point, but {{{x>=4}}} is a requirement in Real Numbers for the domain of f(x).  The far-left point is the one which intersects the x axis, at (4,0).  In other way of saying, {{{f(4)=0}}}.  No other axis intercepts.



Hopefully, a better explanation:
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DOMAIN:  What value are acceptable for x?
Square Root function for real numbers must accept positive values for zero.  Negative values are not acceptable.

{{{f(x)=sqrt(x-4)}}} must have {{{x-4>=0}}} which means {{{x>=0+4}}} or {{{x>=4}}}.
This means, the DOMAIN of f(x) is {{{highlight(x>=4)}}}.


Checking for intercepts, at x=0, what is f(x)?
{{{f(0)=sqrt(0-4)}}}, NOT acceptable because 0 is not in the domain of f(x).  Recall, we just found that the domain of f(x) is {{{x>=4}}}, and {{{0<4}}}, so x cannot be 0.  This means, f(x) will not cross the y-axis.
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What about f(x)=0?  What would x be for this?
{{{f(x)=0=sqrt(x-4)}}}
{{{0=sqrt(x-4)}}}
Square both sides,
{{{0=x-4}}}
{{{x=4}}}
This is the point, (4,0) as the x-intercept.


Your will recognize the equation as half of a parabola with horizontal axis of symmetry:


{{{graph(300,300,-2,12,-2,7,sqrt(x-4))}}}