Find the points at which the graphs of `y=1/2x^2 -3` and y=x +1 intersect.

Options

A. (-2,-3) (4,5)
B. (-1,-2) (5,4)
C. (-4,-3) (-2,4)
D. (-2,-1) (4,5)

Answer

To find the points at which the graphs of \( y = \frac{1}{2}x^2 - 3 \) and \( y = x + 1 \) intersect, we will follow these steps:

### Step 1: Set the equations equal to each other
Since both equations equal \( y \), we can set them equal to find the \( x \) values where they intersect.

\[ \frac{1}{2}x^2 - 3 = x + 1 \]

### Step 2: Rearrange the equation
To solve for \( x \), we need to rearrange the equation into standard quadratic form \( ax^2 + bx + c = 0 \). First, we will move all terms to one side:

\[ \frac{1}{2}x^2 - x - 3 - 1 = 0 \]

This simplifies to:

\[ \frac{1}{2}x^2 - x - 4 = 0 \]

### Step 3: Eliminate the fraction
To eliminate the fraction, multiply the entire equation by 2:

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

### Step 4: Factor the quadratic equation
Now we will factor the quadratic equation \( x^2 - 2x - 8 = 0 \). We need two numbers that multiply to \(-8\) and add to \(-2\). These numbers are \(-4\) and \(2\).

Thus, we can factor the equation as:

\[ (x - 4)(x + 2) = 0 \]

### Step 5: Solve for \( x \)
Setting each factor equal to zero gives us the solutions for \( x \):

\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 6: Find corresponding \( y \) values
Now we will substitute these \( x \) values back into either original equation to find the corresponding \( y \) values. We will use \( y = x + 1 \):

For \( x = 4 \):
\[ y = 4 + 1 = 5 \]

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

### Step 7: Write the intersection points
The points of intersection are:
\[ (4, 5) \quad \text{and} \quad (-2, -1) \]

### Final Answer
The graphs of \( y = \frac{1}{2}x^2 - 3 \) and \( y = x + 1 \) intersect at the points \( (4, 5) \) and \( (-2, -1) \).

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