1. **State the problem:** We are given the function $$f(x) = x^3 + 3x^2 + 5x + 18$$ and need to find all points $$(x,y)$$ on the graph where the tangent line has slope $$5.5$$. 2. **Recall the formula:** The slope of the tangent line to the graph of $$f$$ at $$x$$ is given by the derivative $$f'(x)$$. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x^2) + \frac{d}{dx}(5x) + \frac{d}{dx}(18) = 3x^2 + 6x + 5 + 0 = 3x^2 + 6x + 5$$ 4. **Set the derivative equal to the slope:** $$3x^2 + 6x + 5 = 5.5$$ 5. **Simplify the equation:** $$3x^2 + 6x + 5 - 5.5 = 0$$ $$3x^2 + 6x - 0.5 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=3$$, $$b=6$$, $$c=-0.5$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 6^2 - 4 \times 3 \times (-0.5) = 36 + 6 = 42$$ Calculate the roots: $$x = \frac{-6 \pm \sqrt{42}}{2 \times 3} = \frac{-6 \pm \sqrt{42}}{6}$$ 7. **Find the corresponding $$y$$ values:** Substitute each $$x$$ back into $$f(x)$$: For $$x_1 = \frac{-6 + \sqrt{42}}{6}$$: $$y_1 = \left(\frac{-6 + \sqrt{42}}{6}\right)^3 + 3\left(\frac{-6 + \sqrt{42}}{6}\right)^2 + 5\left(\frac{-6 + \sqrt{42}}{6}\right) + 18$$ For $$x_2 = \frac{-6 - \sqrt{42}}{6}$$: $$y_2 = \left(\frac{-6 - \sqrt{42}}{6}\right)^3 + 3\left(\frac{-6 - \sqrt{42}}{6}\right)^2 + 5\left(\frac{-6 - \sqrt{42}}{6}\right) + 18$$ 8. **Final answer:** The points on the graph where the tangent slope is $$5.5$$ are: $$\left(\frac{-6 + \sqrt{42}}{6}, y_1\right)$$ and $$\left(\frac{-6 - \sqrt{42}}{6}, y_2\right)$$ where $$y_1$$ and $$y_2$$ are calculated as above.