$$\{(x,y):(2x−1)^2+(2y−4)^2=1\}$$
Can anybody help me graph this function and state its domain and range?
My answer for the domain is $[-1.5,1.5]$, and for the range is $[1,3]$, i had already put in consideration of radius $= 1$, but the answer seems to be wrong.
On
The problem with your approach is that since you have $2x$ and $2y$ instead of $x$ and $y$, the centre of the circle is not actually $(1,4)$.
To find the centre, rewrite the equation into standard form $(x-a)^2 + (y-b)^2 = r^2$ by factoring out a $4$:
$$4(x-0.5)^2 + 4(y-2)^2 = 1 \Rightarrow (x-0.5)^2 + (y-2)^2 = 0.25$$
and the rest is straightforward.
You can normalize your equation to $$4(x-\frac 12)^2+4(y-2)^2=1\iff (x-\frac 12)^2+(y-2)^2=\left(\frac 12\right)^2$$
Giving you a circle of center $(\frac 12,2)$ or radius $\frac 12$.
You can then calculate the horizontal and vertical extension, i.e. the rectangular box enclosing the circle is $[0,1]\times[1.5,2.5]$ but I would not really call that domain nor range.
Anyway, if you consider implicit solve $y(x)=2\pm \sqrt{\frac 14-(x-\frac 12)^2}$ then you have $2$ functions $y_1$ and $y_2$ and not one.
The both have the same domain $[0,1]$ but one has range $[2,2.5]$ and the other has range $[1.5,2]$.