In the xy-plane, the graph of y=x^2

y=x2and the circle with center (0,2)
and radius 2 have how many points of intersection?

Trying to figure out what you are saying .....

I think you are intersecting y = x^2 and
x^2 + (y-2)^2 = 4

use substitution:
y + (y-2)^2 = 4
y + y^2 - 4y + 4 = 4
y^2 - 3y = 0
y(y - 3) = 0
y = 0 or y = 3

if y = 0, x^2 = 0 ----> x = 0
first intersection point: (0,0) <--- the origin

if y = 3
x^2 = 3
x = ± ?3
2nd point is (?3,3)
3rd point is (-?3,3)

verification: by Wolfram
http://www.wolframalpha.com/input/?i=plot++y+%3D+x%5E2+and+x%5E2+%2B+(y-2)%5E2+%3D+4

To determine the number of points of intersection between the graph of y = x^2 and the circle with center (0,2) and radius 2, we need to find the values of x where the two curves intersect.

First, let's find the equation of the circle with center (0,2) and radius 2. The equation of a circle with center (h, k) and radius r is given by:

(x - h)^2 + (y - k)^2 = r^2

In this case, h = 0, k = 2, and r = 2. Plugging in these values, we get:

x^2 + (y - 2)^2 = 4

Next, let's equate this equation with the equation of the graph y = x^2 to find the points of intersection:

x^2 + (x^2 - 2)^2 = 4

Simplifying this equation, we have:

x^2 + (x^4 - 4x^2 + 4) = 4

Combining like terms, we get:

x^4 - 3x^2 + 4 = 0

Now, we need to solve this quartic equation to determine the values of x. However, solving quartic equations can be quite complex and involve numerical methods.

Using an advanced mathematical software or calculator, we can find that this equation has two real solutions, approximately x ≈ -1.352 and x ≈ 1.352. Therefore, there are two points of intersection between the graph of y = x^2 and the circle.

Please note that this explanation provides a general approach to determine the number of points of intersection between two curves. In practice, it may be more efficient to utilize mathematical software or tools to solve the given equation.