Let $b$ and $c$ be fixed real numbers and let the ten points $(j,y_j )$ for $j=1,2,\ldots,10$ lie on the parabola $y =x^2 +bx+c.$ For $j=1,2,\ldots, 9$ let $I_j$ be the intersection of the tangents to the given parabola at $(j, y_j )$ and $(j+1, y_{j+1}).$ Determine the poynomial function $y=g(x)$ of least degree whose graph passes through all nine points $I_j .$
Given the parabola \( y = x^2 + bx + c \) and the points \( P(j, y_j) \) and \( Q(j+1, y_{j+1}) \), where \( y_j = j^2 + bj + c \) and \( y_{j+1} = (j+1)^2 + b(j+1) + c \).

1. **Find the equations of the tangents at \( P \) and \( Q \):**

   The derivative of \( y = x^2 + bx + c \) is \( y' = 2x + b \). Therefore, the slope of the tangent at \( P(j, y_j) \) is \( 2j + b \), and the equation of the tangent line at \( P \) is:
   \[
   y - y_j = (2j + b)(x - j)
   \]
   Substituting \( y_j = j^2 + bj + c \):
   \[
   y - (j^2 + bj + c) = (2j + b)(x - j)
   \]
   Simplifying:
   \[
   y = (2j + b)x - (2j + b)j + j^2 + bj + c
   \]
   \[
   y = (2j + b)x - 2j^2 - bj + j^2 + bj + c
   \]
   \[
   y = (2j + b)x - j^2 + c
   \]

   Similarly, the slope of the tangent at \( Q(j+1, y_{j+1}) \) is \( 2(j+1) + b = 2j + 2 + b \), and the equation of the tangent line at \( Q \) is:
   \[
   y - y_{j+1} = (2j + 2 + b)(x - (j+1))
   \]
   Substituting \( y_{j+1} = (j+1)^2 + b(j+1) + c \):
   \[
   y - ((j+1)^2 + b(j+1) + c) = (2j + 2 + b)(x - (j+1))
   \]
   Simplifying:
   \[
   y = (2j + 2 + b)x - (2j + 2 + b)(j+1) + (j+1)^2 + b(j+1) + c
   \]
   \[
   y = (2j + 2 + b)x - (2j + 2 + b)(j+1) + j^2 + 2j + 1 + bj + b + c
   \]
   \[
   y = (2j + 2 + b)x - (2j^2 + 2j + 2j + 2 + bj + b + 2j + 2 + b) + j^2 + 2j + 1 + bj + b + c
   \]
   \[
   y = (2j + 2 + b)x - j^2 - 2j - 1 + c
   \]

2. **Find the intersection \( I_j \) of the tangents:**

   Set the equations of the tangents equal to each other to find the intersection point \( I_j \):
   \[
   (2j + b)x - j^2 + c = (2j + 2 + b)x - (j^2 + 2j + 1) + c
   \]
   Simplifying:
   \[
   (2j + b)x - j^2 + c = (2j + 2 + b)x - j^2 - 2j - 1 + c
   \]
   \[
   (2j + b)x = (2j + 2 + b)x - 2j - 1
   \]
   \[
   0 = 2x - 2j - 1
   \]
   \[
   2x = 2j + 1
   \]
   \[
   x = j + \frac{1}{2}
   \]

   Substitute \( x = j + \frac{1}{2} \) back into one of the tangent equations to find \( y \):
   \[
   y = (2j + b)\left(j + \frac{1}{2}\right) - j^2 + c
   \]
   \[
   y = (2j + b)j + (2j + b)\frac{1}{2} - j^2 + c
   \]
   \[
   y = 2j^2 + bj + j + \frac{b}{2} - j^2 + c
   \]
   \[
   y = j^2 + bj + j + \frac{b}{2} + c
   \]
   \[
   y = j^2 + bj + j + \frac{b}{2} + c
   \]

3. **Eliminate \( j \) to find the polynomial function \( y = g(x) \):**

   Since \( x = j + \frac{1}{2} \), we have \( j = x - \frac{1}{2} \). Substitute \( j = x - \frac{1}{2} \) into the expression for \( y \):
   \[
   y = (x - \frac{1}{2})^2 + b(x - \frac{1}{2}) + (x - \frac{1}{2}) + \frac{b}{2} + c
   \]
   \[
   y = (x^2 - x + \frac{1}{4}) + b(x - \frac{1}{2}) + x - \frac{1}{2} + \frac{b}{2} + c
   \]
   \[
   y = x^2 - x + \frac{1}{4} + bx - \frac{b}{2} + x - \frac{1}{2} + \frac{b}{2} + c
   \]
   \[
   y = x^2 + bx + c - \frac{1}{4}
   \]

Thus, the polynomial function \( y = g(x) \) of least degree whose graph passes through all nine points \( I_j \) is:
\[
y = x^2 + bx + c - \frac{1}{4}
\]

The final answer is \( \boxed{ y = x^2 + bx + c - \frac{1}{4} } \).