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### Problem Breakdown
We need to find the points on the graph of the parabola \( y = x^2 \) that are closest to the point \( (14, \frac{1}{2}) \).

1. **Distance Formula**: The distance \( d \) between any point on the parabola \( (x, x^2) \) and the point \( (14, \frac{1}{2}) \) is given by:
   \[
   d = \sqrt{(x - 14)^2 + \left(x^2 - \frac{1}{2}\right)^2}
   \]

2. **Minimizing \( d \)**: Instead of minimizing \( d \), we can minimize \( d^2 \) (since the square root function is monotonically increasing, minimizing \( d^2 \) will also minimize \( d \)).

   So, let:
   \[
   f(x) = (x - 14)^2 + \left(x^2 - \frac{1}{2}\right)^2
   \]

3. **Derivative of \( f(x) \)**: To find the minimum, we differentiate \( f(x) \) with respect to \( x \) and set the derivative equal to zero.

   Expanding \( f(x) \):
   \[
   f(x) = (x - 14)^2 + \left(x^2 - \frac{1}{2}\right)^2
   \]
   Expanding each term, we get:
   \[
   f(x) = (x - 14)^2 + \left(x^2 - \frac{1}{2}\right)^2 = (x^2 - 28x + 196) + (x^4 - x^2 + \frac{1}{4})
   \]
   \[
   f(x) = x^4 + x^2 - x^2 - 28x + 196 + \frac{1

Find the coordinates of the points on the graph of the parabola y = x^2 that are closest to the point (14, 1/2).

This problem explores finding points on the parabola y = x^2 that are closest to a specific point, (14, 1/2). It involves applying calculus to minimize the distance function, using derivatives and the distance formula.

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