Question 1157204
 {{{f(x)=sqrt(x-2)}}}......... if the {{{f(x)=y}}} is squared, the parabola is horizontal (opens left or right),

so, the leftmost point is a vertex

rewrite in vertex form:

 {{{y=sqrt(x-2)}}}
{{{y^2=(x-2)}}} => compare to {{{(y-k)^2=a(x-2)}}}, => {{{h=2}}} and {{{k=0}}}=>the vertex is at ({{{2}}}, {{{0}}})

make a table and use equation to find {{{3}}} more points:

{{{x}}}|{{{y}}}
{{{2}}}, {{{0}}}
{{{3}}}, {{{1}}}.....{{{y=sqrt(3-2)=sqrt(1)}}}=> {{{y=1}}} or {{{y=-1}}}
{{{3}}}, {{{-1}}}
{{{4}}}, {{{1.4}}}.....{{{y=sqrt(4-2)=2}}}=> {{{y=1.4}}} or {{{y=-1.4}}}
{{{4}}}, {{{-1.4}}}
{{{6}}}, {{{2}}}.....{{{y=sqrt(6-2)=sqrt(4)}}}=> {{{y=2}}} or {{{y=-2}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,0,.12), locate(2,0,v(2,0)),
circle(3,1,.12), locate(3,1,p(3,1)),
circle(3,-1,.12), locate(3,-1,p(3,-1)),
circle(4,1.4,.12), locate(4,1.4,p(4,1.4)),
circle(4,-1.4,.12), locate(4,-1.4,p(4,-1.4)),
circle(6,2,.12), locate(6,2,p(6,2)),
circle(6,-2,.12), locate(6,-2,p(6,-2)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(x-2), -sqrt(x-2))) }}}