Question 939711
domain:

{{{4(x-1)^2+16y^2=32}}}

{{{16y^2=32-4(x-1)^2}}}

{{{y^2=32/16-4(x-1)^2/16}}}

{{{y^2=2-(1/4)(x-1)^2}}}

{{{y=sqrt(2-(1/4)(x-1)^2)}}} ; to have real solution, determinante

{{{2-(1/4)(x-1)^2}}} must be greater or equal to zero

so,
{{{2-(1/4)(x-1)^2>=0}}} ...solve for {{{x}}}

{{{2>=(1/4)(x-1)^2}}}

{{{8>=(x-1)^2}}}

{{{sqrt(8)>=x-1}}}

{{{1+sqrt(2*4)>=x}}}

{{{1+2sqrt(2)>=x}}}...since for {{{sqrt(2)}}} we have positive and negative solution,

means, solutions are {{{1+2sqrt(2)>=x}}} or {{{1-2sqrt(2)<=x}}}

domain is:

{  {{{x}}} element {{{R}}} : {{{1-2sqrt(2)<=x<=1+2sqrt(2)}}}  }
(assuming a function from reals to reals)

to find the range, find out what is {{{y}}} if {{{x}}} goes from zero to infinity

{{{y=sqrt(2-(1/4)(x-1)^2)}}}

{{{x}}} goes greater, and the smallest value of{{{(1/4)(x-1)^2)}}} could be  zero, then {{{y=sqrt(2)}}}

so, the range is:

{  {{{y}}} element {{{R}}} : {{{0<=y<=sqrt(2)}}}  }