If (x,y) represents a point on the the graph of `y=2x+1`, which of the following could be a portion of the graph of the set of points `(x,y^(2))`?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to find the graph of the set of points \((x, y^2)\) given that \(y = 2x + 1\).

### Step-by-Step Solution:

1. **Start with the equation of the line**:
\[ y = 2x + 1 \]

2. **Substitute \(y\) into \(y^2\)**:
We want to express \(y^2\) in terms of \(x\):
\[ y^2 = (2x + 1)^2 \]

3. **Expand the expression**:
Now, expand the squared term:
\[ y^2 = (2x + 1)(2x + 1) = 4x^2 + 4x + 1 \]

4. **Rearrange the equation**:
We can write this as:
\[ y^2 = 4x^2 + 4x + 1 \]

5. **Identify the standard form of a parabola**:
The equation \(y^2 = 4ax\) is the standard form of a parabola that opens to the right. To convert our equation into a similar form, we can rearrange it:
\[ y^2 = 4(x^2 + x + \frac{1}{4}) \]

6. **Complete the square**:
To complete the square for \(x^2 + x\):
\[ x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4} \]
Thus, we can rewrite the equation as:
\[ y^2 = 4\left((x + \frac{1}{2})^2 - \frac{1}{4} + \frac{1}{4}\right) = 4\left(x + \frac{1}{2}\right)^2 \]

7. **Final form**:
Therefore, we have:
\[ y^2 = 4\left(x + \frac{1}{2}\right)^2 \]
This shows that the graph is a parabola that opens to the right with its vertex at \((- \frac{1}{2}, 0)\).

### Conclusion:
The graph of the set of points \((x, y^2)\) is a vertical parabola that opens to the right, with its vertex at the point \((- \frac{1}{2}, 0)\).