find the points of intersection of the following pairs of equations.

a. y=x^2 and y=x +2

b. y=x^2 and y=8-x^2

c. y=x^2 and x=y^2

a. Substitute (x + 2) from the second eq., simplify and solve for x. Then solve for y.

The other two equations are done the same way.

a. x+2 = x^2
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0

can you finish it?

for for like a. it would be (2, 4) and (-1, 1)?

Yes. Sorry it took so long to reply.

To find the points of intersection of two equations, you need to find the values of x and y that satisfy both equations simultaneously. This means that the x and y values of the points of intersection will be the solutions to both equations.

a. y=x^2 and y=x+2:

To find the points of intersection, set the equations equal to each other:
x^2 = x + 2

Rearrange the equation to form a quadratic equation:
x^2 - x - 2 = 0

Now, you can either factorize or use the quadratic formula to solve for x. In this example, the quadratic equation can be factored:
(x - 2)(x + 1) = 0

Set each factor equal to zero and solve:
x - 2 = 0 or x + 1 = 0

Solving each equation gives us two potential x-values for the points of intersection:
x = 2 or x = -1

Substitute these x-values into either of the original equations to find the corresponding y-values:
If x = 2, then y = (2)^2 = 4.
If x = -1, then y = (-1)^2 + 2 = 3.

Therefore, the points of intersection are (2, 4) and (-1, 3).

b. y=x^2 and y=8-x^2:

Set the equations equal to each other:
x^2 = 8 - x^2

Rearrange the equation:
2x^2 = 8

Divide both sides by 2:
x^2 = 4

Take the square root of both sides (considering both positive and negative roots):
x = ±2

Substitute these x-values into either of the original equations to find the corresponding y-values:
If x = 2, then y = (2)^2 = 4.
If x = -2, then y = (-2)^2 = 4.

Therefore, the points of intersection are (2, 4) and (-2, 4).

c. y=x^2 and x=y^2:

Set the equations equal to each other:
x^2 = y^2

Take the square root of both sides (considering both positive and negative roots):
x = ±y

Substitute x = y into either of the original equations:
If y = x, then y = (y)^2

This gives us a quadratic equation, y^2 - y = 0, which can be factored as y(y - 1) = 0.

Set each factor equal to zero and solve:
y = 0 or y - 1 = 0

Solving each equation gives us two potential y-values for the points of intersection:
y = 0 or y = 1

Substitute these y-values back into the original equation to find the corresponding x-values:
If y = 0, then x = (0)^2 = 0.
If y = 1, then x = (1)^2 = 1.

Therefore, the points of intersection are (0, 0) and (1, 1).