Question 1091208
the radical function would be:

{{{f(x)=a*sqrt(x-b)+c}}}

use the points 
({{{1}}},{{{-2}}}),
 ({{{5}}},{{{0}}}),
({{{10}}},{{{1}}}) to find {{{a}}},{{{b}}}, and {{{c}}}

{{{f(x)=a*sqrt(x-b)+c}}}...........({{{1}}},{{{-2}}})

{{{-2=a*sqrt(1-b)+c}}}....solve for {{{c}}}


{{{c=-2-a*sqrt(1-b)}}}.......eq.1

{{{f(x)=a*sqrt(x-b)+c}}}..........({{{5}}},{{{0}}})
{{{0=a*sqrt(5-b)+c}}}..........solve for {{{c}}}
{{{c=-a*sqrt(5-b)}}}.......eq.2


{{{f(x)=a*sqrt(x-b)+c}}}..........({{{10}}},{{{1}}})
{{{1=a*sqrt(10-b)+c}}}..........solve for {{{c}}}

{{{c=1-a*sqrt(10-b)}}}.......eq.3

from eq.1 and eq.2 we have:

{{{-2-a*sqrt(1-b)=-a*sqrt(5-b)}}}.....both sides multiply by  {{{-1}}}

{{{2+a*sqrt(1-b)=a*sqrt(5-b)}}}............square both sides

{{{(2+a*sqrt(1-b))^2=(a*sqrt(5-b))^2}}}

{{{4+2a*sqrt(1-b)+a^2(1-b)=a^2(5-b)}}}

{{{2a*sqrt(1-b)=a^2(5-b)-a^2(1-b)-4}}}
{{{2a*sqrt(1-b)=5a^2-a^2b-a^2+a^2b-4}}}
{{{2a*sqrt(1-b)=4a^2-4}}}
{{{sqrt(1-b)=4a^2/2a-4/2a}}}
{{{sqrt(1-b)=2a-2/a}}}
{{{sqrt(1 - b) = (2 (a - 1) (a + 1))/a}}}.......square both sides
{{{1 - b = ((2 (a - 1) (a + 1))/a)^2}}}....solve for {{{b}}}
{{{b=1 - (4 (a - 1)^2 (a + 1)^2)/a^2}}}..............eq.b

from eq.2 and eq.3 we have:

{{{1-a*sqrt(10-b)=-a*sqrt(5-b)}}}
{{{1-asqrt(10-b)+asqrt(5-b)=0}}}

{{{1+a(sqrt(5-b)-sqrt(10-b))=0}}}..........solve for {{{b}}}

{{{b = (-25 a^4 + 30 a^2 - 1)/(4 a^2)}}}...........eq.b1

from .eq.b and .eq.b1:

{{{1 - (4 (a - 1)^2 (a + 1)^2)/a^2= (-25 a^4 + 30 a^2 - 1)/(4 a^2)}}} ......solve for {{{a}}}

{{{-((2 a^2 - a - 2) (2 a^2 + a - 2))/a^2 = -(25 a^4 - 30 a^2 + 1)/(4 a^2)}}}

to make it short, you will have 
{{{highlight(a=1)}}} or {{{highlight(a=-1)}}}


go to {{{b=1 - (4 (a - 1)^2 (a + 1)^2)/a^2}}}..............eq.b, plug in {{{a=1}}} 

{{{b=1 - (4 (1 - 1)^2 (1+ 1)^2)/1^2}}}
{{{b=1 - (4 *0 *4)/1}}}
{{{highlight(b=1 )}}}

{{{b=1 - (4 (a - 1)^2 (a + 1)^2)/a^2}}}..............eq.b, plug in {{{a=-1}}} 

{{{b=1 - (4 (-1 - 1)^2 (-1+ 1)^2)/(-1)^2}}}
{{{b=1 - (4 *4 *0)/1}}}
{{{highlight(b=1 )}}}; so, there is one double solution for {{{b}}}



now find {{{c}}}:

{{{c=-a*sqrt(5-b)}}}.......eq.2...............{{{highlight(b=1)}}} and {{{highlight(a=1)}}}
{{{c=-1*sqrt(5-1)}}}
{{{c=-sqrt(4)}}}
{{{highlight(c=-2)}}} 

{{{c=-a*sqrt(5-b)}}}.......eq.2...............{{{highlight(b=1)}}} and {{{highlight(a=-1)}}}
{{{c=-(-1)*sqrt(5-1)}}}
{{{c=1sqrt(4)}}}

{{{highlight(c=2)}}} 

your solutions are:

{{{a=1}}} , {{{b=1}}},  {{{c=-2}}}  or

{{{a=-1}}} , {{{b=1}}},  {{{c=2}}} 

the radical function would be:

{{{f(x)=a*sqrt(x-b)+c}}} .....if {{{a=1}}} , {{{b=1}}},  {{{c=-2}}}
{{{highlight(f(x)=sqrt(x-1)-2)}}}

or
{{{f(x)=a*sqrt(x-b)+c}}} .....if {{{a=-1}}} , {{{b=1}}},  {{{c=2}}} 
{{{f(x)=-sqrt(x-1)+2}}}-> 
since given point  ({{{1}}},{{{-2}}}), the value of {{{x=1}}} will give us {{{sqrt(1-1)=sqrt(0)}}} and {{{y=2}}}
so, disregard {{{f(x)=-sqrt(x-1)+2}}}


{{{drawing( 600, 600, -10, 15, -10, 10, 
circle(1,-2,.12),circle(5,0,.12),circle(10,1,.12),
locate(1,-2,p(1,-2)),locate(5,-0.4,p(5,0)),locate(10,1,p(10,1)),
graph( 600, 600, -10, 15, -10, 10, sqrt(x-1)-2)) }}}