Step 1 :Given the function \(y=\frac{1}{3} x^{3}-4 x^{2}+20 x+9\), we are asked to find the point(s) on the graph at which the tangent line has slope 5.
Step 2 :The slope of the tangent line to the graph of a function at a particular point is given by the derivative of the function at that point.
Step 3 :First, we find the derivative of the function, which is \(y'=x^{2}-8x+20\).
Step 4 :Next, we set the derivative equal to 5 and solve for x, which gives us the x-coordinates of the points where the slope of the tangent line is 5. The solutions are \(x=3\) and \(x=5\).
Step 5 :Finally, we substitute these x-coordinates back into the original function to find the corresponding y-coordinates. This gives us the points \((3, 42)\) and \((5, 50.67)\).
Step 6 :Final Answer: The points on the graph at which the tangent line has slope 5 are \(\boxed{(3, 42)}\) and \(\boxed{(5, 50.67)}\).