Find an expression for the function whose graph is the given curve.
The top half of the circle x^2 + (y-2)^2 = 4
Please Help. I don't even know where to start.
Hints:
1. Solve for y in terms of x.
2. You will have an expression involving a square-root of an expression.
3. You will need to restrict the expression under the radical to be non-negative, or else you will end up with a complex quantity.
4. For the top half of the curve (circle), restrict the value of the square-root to be non-negative, i.e. take the positive value of the square-root only.
So what do you mean by y in terms of x? I solved x^2 + (y-2)^2 = 4 and I got y^2 - 4y + 8 = x^2 ?
I would isolate y on the left like:
(y-2)²=4-x²
and take square-root on both sides.
transpose the -2 to the right and you'll get your f(x)=y=....
To find an expression for the function whose graph is the top half of the circle, we need to isolate the variable "y" in the equation of the circle.
The equation of the circle is given as:
x^2 + (y-2)^2 = 4
To isolate the variable "y," we start by subtracting x^2 from both sides of the equation:
(y-2)^2 = 4 - x^2
Next, we take the square root of both sides of the equation:
√((y-2)^2) = ± √(4 - x^2)
Simplifying:
y - 2 = ± √(4 - x^2)
Finally, adding 2 to both sides of the equation gives us the expression for the function:
y = 2 ± √(4 - x^2)
Therefore, the expression for the function whose graph is the top half of the circle x^2 + (y-2)^2 = 4 is y = 2 ± √(4 - x^2).