1. **State the problem:** Find the equation of the tangent line to the curve $y = x\sqrt{x}$ that is parallel to the line $y = 1 + 3x$. 2. **Identify the slope of the given line:** The line $y = 1 + 3x$ has slope $m = 3$. 3. **Find the derivative of the curve to get the slope of the tangent line:** Rewrite the function as $y = x \cdot x^{1/2} = x^{3/2}$. Using the power rule, the derivative is: $$y' = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x}$$ 4. **Set the derivative equal to the slope of the given line to find the point(s) of tangency:** $$\frac{3}{2} \sqrt{x} = 3$$ Divide both sides by 3: $$\frac{1}{2} \sqrt{x} = 1$$ Multiply both sides by 2: $$\sqrt{x} = 2$$ Square both sides: $$x = 4$$ 5. **Find the corresponding $y$ value on the curve:** $$y = x \sqrt{x} = 4 \times 2 = 8$$ 6. **Write the equation of the tangent line using point-slope form:** $$y - y_1 = m (x - x_1)$$ Substitute $m=3$, $x_1=4$, $y_1=8$: $$y - 8 = 3 (x - 4)$$ Simplify: $$y = 3x - 12 + 8 = 3x - 4$$ **Final answer:** The equation of the tangent line is $y = 3x - 4$.