Find an expression for the function whose graph is the given curve.
The top half of the circle x^2 + (y-2)^2 = 4
My final answer is y= 2 +/- sqrt of x-4
However, I circled the positive sign because that's the top half. Is this right? I was not sure whether it was sqrt of x-4 or sqrt of x^2-4
you are solving for y, so ...
(y-2)^2 = 4 - x^2
y - 2 = ± √(4-x^2)
since you want the top part
y = + √(4-x^2) + 2
incidentally, √(4-x^2) ≠ 2-x
e.g. √(16 - 4) = √12 ≠ 4-2
To find the equation for the function whose graph is the top half of the circle x^2 + (y-2)^2 = 4, you need to isolate y in terms of x.
1. Start with the equation of the circle: x^2 + (y-2)^2 = 4.
2. Expand the equation: x^2 + y^2 - 4y + 4 = 4.
3. Subtract 4 from both sides to isolate the y terms: x^2 + y^2 - 4y = 0.
4. Rewrite the equation: y^2 - 4y = -x^2.
5. Complete the square on the left side of the equation: y^2 - 4y + 4 = -x^2 + 4.
6. Rewrite the equation: (y-2)^2 = -x^2 + 4.
7. Take the square root of both sides to solve for y: y - 2 = ±√(-x^2 + 4).
8. Add 2 to both sides: y = 2 ± √(-x^2 + 4).
Now let's analyze the expression. Since we want the top half of the circle, we only need the positive square root. Therefore, the expression for the function is:
y = 2 + √(-x^2 + 4).
Note that the expression is relating y to x through the equation only, and there is no need to subtract 4 inside the square root.