Question

Create two functions and find their intersection points. What two functions can be used to solve the following equation, and what is their solution?

Answer 1

To find the intersection points of two functions, we can choose a simple pair of functions for illustration. Let's consider the following two functions:

1. \( f(x) = x^2 - 4 \)
2. \( g(x) = 2x - 3 \)

### Step 1: Set the equations equal to each other.

To find the intersection points, set \( f(x) = g(x) \):

\[
x^2 - 4 = 2x - 3
\]

### Step 2: Rearrange the equation.

Rearranging gives us:

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

### Step 3: Solve the quadratic equation.

We can use the quadratic formula to solve for \( x \):

The quadratic formula is:

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

For our equation \( x^2 - 2x - 1 = 0 \):

- \( a = 1 \)
- \( b = -2 \)
- \( c = -1 \)

Plugging in these values:

\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}
\]
\[
x = \frac{2 \pm \sqrt{4 + 4}}{2}
\]
\[
x = \frac{2 \pm \sqrt{8}}{2}
\]
\[
x = \frac{2 \pm 2\sqrt{2}}{2}
\]
\[
x = 1 \pm \sqrt{2}
\]

### Step 4: Find the y-coordinates.

Now, we need to find the corresponding \( y \) values using either function. We will use \( g(x) = 2x - 3 \):

1. For \( x = 1 + \sqrt{2} \):

\[
g(1 + \sqrt{2}) = 2(1 + \sqrt{2}) - 3 = 2 + 2\sqrt{2} - 3 = -1 + 2\sqrt{2}
\]

2. For \( x = 1 - \sqrt{2} \):

\[
g(1 - \sqrt{2}) = 2(1 - \sqrt{2}) - 3 = 2 - 2\sqrt{2} - 3 = -1 - 2\sqrt{2}
\]

### Conclusion: Intersection Points.

The intersection points of the functions are:

1. \( (1 + \sqrt{2}, -1 + 2\sqrt{2}) \)
2. \( (1 - \sqrt{2}, -1 - 2\sqrt{2}) \)

Thus, the two functions \( f(x) = x^2 - 4 \) and \( g(x) = 2x - 3 \) intersect at these two points.