When the solutions to each of the two equations below are graphed in the xy coordinate plane, the graphs of the solutions intersect at two places. write the y-coordinate of the points of intersection in the boxes below in order from smallest to largest. y=2x and y=x^2-3

The points of intersection are at (3, 6) and (-1, -2).

Since both of these equations have y isolated, we can set them equal to each other:

2x=x²-3

We want all of the variables on one side, so subtract 2x:
2x-2x = x²-3-2x
0=x²-3-2x

Write the quadratic in standard form:
0=x²-2x-3

This is easily factorable, as there are factors of -3 that will sum to -2.  -3(1)=-3 and -3+1=-2:
0=(x-3)(x+1)

Using the zero product property we know that either x-3=0 or x+1=0; therefore x=3 or x=-1.

Substituting this into the first equation (it is simpler):
y=2(3) = 6
y=2(-1)=-2

Therefore the coordinates are (3, 6) and (-1, -2).

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