find an expression for the function whose graph consists of the line segment from the point (2,2) to the point (-1,0) together with the top half of the circle with center the origin and radius 1
To find an expression for the given function, we need to describe it in two separate parts: the line segment and the top half of the circle.
First, let's consider the line segment. We know that the line passes through the points (2, 2) and (-1, 0). To determine the equation of a line, we usually use the slope-intercept form: y = mx + b, where m is the slope of the line and b is the y-intercept.
Using the two points, we can find the slope of the line:
slope (m) = (y₂ - y₁) / (x₂ - x₁)
= (0 - 2) / (-1 - 2)
= -2 / -3
= 2/3
Now we have the slope (m), and to find the y-intercept (b), we can substitute the coordinates of one of the points into the equation:
y = mx + b
2 = (2/3)(2) + b
2 = 4/3 + b
To isolate b, we subtract 4/3 from both sides:
2 - 4/3 = b
6/3 - 4/3 = b
2/3 = b
So the equation of our line segment becomes:
y = (2/3)x + 2/3
Now let's consider the top half of the circle. The equation for a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
In this case, the center of the circle is the origin (0, 0), and the radius is 1. Therefore, the equation of the circle is:
x^2 + y^2 = 1^2
x^2 + y^2 = 1
However, we only want the top half of the circle. This means the y-values should be positive or zero.
Combining the equation of the line segment and the top half of the circle, we have:
y = (2/3)x + 2/3 (for the line segment)
x^2 + y^2 = 1 (for the top half of the circle, where y >= 0)
Therefore, the expression for the function whose graph consists of the line segment from (2, 2) to (-1, 0) together with the top half of the circle with center the origin and radius 1 is:
y = (2/3)x + 2/3 (for -1 ≤ x ≤ 2)
x^2 + y^2 = 1 (for -1 ≤ x ≤ 1 and y ≥ 0)