Step 1 :The function given is \(y=x^{3}-4 x^{2}+5 x+3\).
Step 2 :The tangent line to a function at a given point is horizontal if and only if the derivative of the function at that point is zero.
Step 3 :First, we need to find the derivative of the function, which is \(y' = 3x^{2} - 8x + 5\).
Step 4 :Next, we set the derivative equal to zero and solve for x, which gives us the x-values \(x = 1\) and \(x = \frac{5}{3}\).
Step 5 :Finally, we substitute these x-values into the original function to find the corresponding y-values, which gives us the points \((1, 5)\) and \(\left(\frac{5}{3}, \frac{131}{27}\right)\).
Step 6 :Thus, the points on the graph at which the tangent line is horizontal are \(\boxed{(1, 5)}\) and \(\boxed{\left(\frac{5}{3}, \frac{131}{27}\right)}\).