Y=x^2-4 y=x-8

To find the points of intersection between the two equations \( y = x^2 - 4 \) and \( y = x - 8 \), we need to set them equal to each other:

\[ x^2 - 4 = x - 8 \]

Now, let's rearrange the equation:

\[ x^2 - x - 4 + 8 = 0 \]

Simplifying, we get:

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

Now, let's apply the quadratic formula to solve for \( x \). The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \( a = 1 \), \( b = -1 \), and \( c = 4 \). Plugging these values into the formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \] \[ x = \frac{1 \pm \sqrt{1 - 16}}{2} \] \[ x = \frac{1 \pm \sqrt{-15}}{2} \]

Since we have a negative number under the square root, this means the solutions for \( x \) are complex:

\[ x = \frac{1 \pm i\sqrt{15}}{2} \]

Now, we can calculate the corresponding \( y \) values by plugging the \( x \) values back into either original equation. Let’s use \( y = x - 8 \).

For \( x = \frac{1 + i\sqrt{15}}{2} \):

\[ y = \frac{1 + i\sqrt{15}}{2} - 8 = \frac{1 + i\sqrt{15}}{2} - \frac{16}{2} = \frac{-15 + i\sqrt{15}}{2} \]

For \( x = \frac{1 - i\sqrt{15}}{2} \):

\[ y = \frac{1 - i\sqrt{15}}{2} - 8 = \frac{1 - i\sqrt{15}}{2} - \frac{16}{2} = \frac{-15 - i\sqrt{15}}{2} \]

Thus, the points of intersection in complex form are:

1. \(\left(\frac{1 + i\sqrt{15}}{2}, \frac{-15 + i\sqrt{15}}{2}\right)\)
2. \(\left(\frac{1 - i\sqrt{15}}{2}, \frac{-15 - i\sqrt{15}}{2}\right)\)

These points indicate that the curves do not intersect on the real plane; they only intersect in the complex number space.