1. y=x²-4x-2

It seems you are providing equations to analyze. Let's go through them one by one to find their intersections and any other relevant information.

1. \( y = x^2 - 4x - 2 \) and \( y = 2 \)

To find the points of intersection, set the two equations equal to each other:

\[ x^2 - 4x - 2 = 2 \]

Now, simplify:

\[ x^2 - 4x - 4 = 0 \]

Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, c = -4 \):

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)} \] \[ x = \frac{4 \pm \sqrt{16 + 16}}{2} \] \[ x = \frac{4 \pm \sqrt{32}}{2} \] \[ x = \frac{4 \pm 4\sqrt{2}}{2} \] \[ x = 2 \pm 2\sqrt{2} \]

Now we can find the y-values:

Substituting back into \( y = 2 \):

• The points of intersection are:
  • \( (2 + 2\sqrt{2}, 2) \)
  • \( (2 - 2\sqrt{2}, 2) \)

2. \( y = x^2 + 1 \) and \( y = -2x^2 + 4 \)

Setting the two equations equal to each other to find points of intersection:

\[ x^2 + 1 = -2x^2 + 4 \]

Rearranging gives:

\[ 3x^2 - 3 = 0 \]

Factoring out:

\[ 3(x^2 - 1) = 0 \]

So,

\[ x^2 - 1 = 0 \] \[ x^2 = 1 \]

Thus \( x = \pm 1 \).

Now finding y-values for these x-values:

1. For \( x = 1 \): \[ y = 1^2 + 1 = 2 \]

2. For \( x = -1 \): \[ y = (-1)^2 + 1 = 2 \]

Thus the points of intersection for the second pair are:

• \( (1, 2) \)
• \( (-1, 2) \)

Summary

1. The points of intersection for \( y = x^2 - 4x - 2 \) and \( y = 2 \) are:

   • \( (2 + 2\sqrt{2}, 2) \)
   • \( (2 - 2\sqrt{2}, 2) \)

2. The points of intersection for \( y = x^2 + 1 \) and \( y = -2x^2 + 4 \) are:

   • \( (1, 2) \)
   • \( (-1, 2) \)