SOLUTION: Hey! I am working on finding the domain and range of a relation that is not a function. The relation is: 4(x-1)^2+16y^2=32. For domain, I have tried solving for y and got y^2=[-(x

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hey! I am working on finding the domain and range of a relation that is not a function. The relation is: 4(x-1)^2+16y^2=32. For domain, I have tried solving for y and got y^2=[-(x      Log On


   

Question 939711: Hey! I am working on finding the domain and range of a relation that is not a function. The relation is: 4(x-1)^2+16y^2=32.
For domain, I have tried solving for y and got y^2=[-(x-1)^2]/(4)+2. The next step would be square rooting to get the equation under the square radical and taking that equation to put in an inequality. The problem is that the equation is negative, making it undefined over reals. The same problem occurs when solving for the range. Please help me, thanks.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
domain:
4%28x-1%29%5E2%2B16y%5E2=32
16y%5E2=32-4%28x-1%29%5E2
y%5E2=32%2F16-4%28x-1%29%5E2%2F16
y%5E2=2-%281%2F4%29%28x-1%29%5E2
y=sqrt%282-%281%2F4%29%28x-1%29%5E2%29 ; to have real solution, determinante
2-%281%2F4%29%28x-1%29%5E2 must be greater or equal to zero
so,
2-%281%2F4%29%28x-1%29%5E2%3E=0 ...solve for x
2%3E=%281%2F4%29%28x-1%29%5E2
8%3E=%28x-1%29%5E2
sqrt%288%29%3E=x-1
1%2Bsqrt%282%2A4%29%3E=x
1%2B2sqrt%282%29%3E=x...since for sqrt%282%29 we have positive and negative solution,
means, solutions are 1%2B2sqrt%282%29%3E=x or 1-2sqrt%282%29%3C=x
domain is:
{ x element R : 1-2sqrt%282%29%3C=x%3C=1%2B2sqrt%282%29 }
(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%282-%281%2F4%29%28x-1%29%5E2%29
x goes greater, and the smallest value of%281%2F4%29%28x-1%29%5E2%29 could be zero, then y=sqrt%282%29
so, the range is:
{ y element R : 0%3C=y%3C=sqrt%282%29 }