Find an expression for the function whose graph is the given curve.
a) The bottom half of the parabola x+(y-1)^2=0
b) The top half of the circle (x-1)^2 + y^2= 0
I think I have a vague idea of what I need to have in my expression for b; I know it's a circle with a center (1,0) and a radius of 1, so the range is [0,2] and the range [that we're concerned with] is [0, 1]. I think I might need to use the absolute value of x in there somewhere, but other than that I don't know what to do. For both a) and b), I don't really know how to formulate an expression for a function where the slope is not constant. Any help would be appreciated.
a) To find an expression for the function whose graph is the bottom half of the parabola x + (y - 1)^2 = 0, we need to isolate y and solve for it.
x + (y - 1)^2 = 0
(y - 1)^2 = -x
y - 1 = ±√(-x)
To consider the bottom half of the parabola, we only need the negative square root:
y - 1 = -√(-x)
y = 1 - √(-x)
So, the expression for the function whose graph is the bottom half of the parabola is y = 1 - √(-x).
b) To find an expression for the function whose graph is the top half of the circle (x - 1)^2 + y^2 = 0, we need to isolate y and solve for it.
(x - 1)^2 + y^2 = 0
y^2 = - (x - 1)^2
y = ±√(-(x - 1)^2)
To consider the top half of the circle, we only need the positive square root:
y = √(-(x - 1)^2)
However, this expression gives the entire upper half of the circle. To limit it only to the range [0, 1], we can add a restriction to the expression. We know that the circle has center (1,0) and radius 1. On the interval [0, 1] the y-coordinate is always positive, so we can modify the expression to:
y = |√(-(x - 1)^2)|
This accounts for the restriction of the range [0, 1] and gives the expression for the function whose graph is the top half of the circle.
Note: The expression involves the absolute value of √(-(x - 1)^2) because the negative square root will only be valid for x-values outside the range [0, 1].