In the xy-plane, which of the following is a point of intersection between the graphs of `y=x+2 and y=x^(2)+x-2`?

Options

A. `(0, -2)`
B. `(0, 2)`
C. `(1, 0)`
D. `(2, 4)`

Answer

To find the point of intersection between the graphs of the equations \( y = x + 2 \) and \( y = x^2 + x - 2 \), we can follow these steps:

### Step 1: Set the equations equal to each other
Since both equations are equal to \( y \), we can set them equal to find the points of intersection:

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

### Step 2: Simplify the equation
To simplify, we can subtract \( x + 2 \) from both sides:

\[ 0 = x^2 + x - 2 - (x + 2) \]

This simplifies to:

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

### Step 3: Factor the quadratic equation
Now, we can factor the quadratic equation:

\[ 0 = (x - 2)(x + 2) \]

### Step 4: Solve for \( x \)
Setting each factor equal to zero gives us the possible \( x \)-values:

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 5: Find corresponding \( y \)-values
Now, we need to find the corresponding \( y \)-values for each \( x \) using the equation \( y = x + 2 \).

1. For \( x = 2 \):
\[ y = 2 + 2 = 4 \]
So, one point of intersection is \( (2, 4) \).

2. For \( x = -2 \):
\[ y = -2 + 2 = 0 \]
So, another point of intersection is \( (-2, 0) \).

### Conclusion
The points of intersection between the graphs of \( y = x + 2 \) and \( y = x^2 + x - 2 \) are \( (2, 4) \) and \( (-2, 0) \). Among the options provided, \( (2, 4) \) is one of the choices.

### Final Answer
The point of intersection is \( (2, 4) \).

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