Question

1. y=x²-4x-2
y=x=2


2. y=x²+1
y=-2x²+4

Answer 1

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:
- \( (1, 2) \)
- \( (-1, 2) \)