Example 6 i) Sketch the curve \( y=(x-3)^{2} \) and the line \( y=x-1 \) on the same axis. ii) Calculate the area of the finite region enclosed by the curve \( y=(x-3)^{2} \) and the line \( y=x-1 \).

Certainly! Let's tackle **Example 6** step by step. ## Part i) Sketching the Curves ### 1. **Parabola: \( y = (x - 3)^2 \)** - **Vertex:** The parabola \( y = (x - 3)^2 \) opens upwards with its vertex at \( (3, 0) \). - **Symmetry:** It is symmetric about the vertical line \( x = 3 \). - **Key Points:** - When \( x = 2 \), \( y = (2-3)^2 = 1 \) → Point \( (2, 1) \) - When \( x = 4 \), \( y = (4-3)^2 = 1 \) → Point \( (4, 1) \) - When \( x = 1 \), \( y = (1-3)^2 = 4 \) → Point \( (1, 4) \) - When \( x = 5 \), \( y = (5-3)^2 = 4 \) → Point \( (5, 4) \) ### 2. **Line: \( y = x - 1 \)** - **Slope:** The line has a slope of 1. - **Y-intercept:** It crosses the y-axis at \( (0, -1) \). - **Key Points:** - When \( x = 1 \), \( y = 0 \) → Point \( (1, 0) \) - When \( x = 2 \), \( y = 1 \) → Point \( (2, 1) \) - When \( x = 3 \), \( y = 2 \) → Point \( (3, 2) \) - When \( x = 5 \), \( y = 4 \) → Point \( (5, 4) \) ### 3. **Intersection Points** To find where the two curves intersect: \[ (x - 3)^2 = x - 1 \\ x^2 - 6x + 9 = x - 1 \\ x^2 - 7x + 10 = 0 \\ (x - 2)(x - 5) = 0 \\ x = 2 \quad \text{or} \quad x = 5 \] - **Intersection Points:** \( (2, 1) \) and \( (5, 4) \) ### 4. **Plotting the Graphs** Using the key points and intersection points, plot both the parabola and the line on the same coordinate system. The parabola will open upwards with its vertex at \( (3, 0) \), and the line will pass through \( (0, -1) \) and intersect the parabola at \( (2, 1) \) and \( (5, 4) \). **[Since I can't provide a visual sketch here, please graph the points and curves as described to visualize the intersection.]** ## Part ii) Calculating the Area of the Enclosed Region ### 1. **Setting Up the Integral** The area between the two curves from \( x = 2 \) to \( x = 5 \) is given by: \[ \text{Area} = \int_{2}^{5} \left[ \text{Top Function} - \text{Bottom Function} \right] dx \] From the sketch and intersection analysis, the **line \( y = x - 1 \)** is above the **parabola \( y = (x - 3)^2 \)** in this interval. ### 2. **Expressing the Difference of Functions** \[ \text{Top Function} - \text{Bottom Function} = (x - 1) - (x - 3)^2 \] Simplify the expression: \[ (x - 1) - (x^2 - 6x + 9) = -x^2 + 7x - 10 \] ### 3. **Integrating** \[ \text{Area} = \int_{2}^{5} (-x^2 + 7x - 10) \, dx \] Find the antiderivative: \[ \int (-x^2 + 7x - 10) \, dx = -\frac{x^3}{3} + \frac{7x^2}{2} - 10x \] Evaluate from \( x = 2 \) to \( x = 5 \): \[ \left[ -\frac{5^3}{3} + \frac{7 \cdot 5^2}{2} - 10 \cdot 5 \right] - \left[ -\frac{2^3}{3} + \frac{7 \cdot 2^2}{2} - 10 \cdot 2 \right] \] Compute each part: \[ \left[ -\frac{125}{3} + \frac{175}{2} - 50 \right] - \left[ -\frac{8}{3} + 14 - 20 \right] \] Convert to a common denominator (6): \[ \left[ -\frac{250}{6} + \frac{525}{6} - \frac{300}{6} \right] - \left[ -\frac{16}{6} + \frac{84}{6} - \frac{120}{6} \right] \] Simplify: \[ \left[ \frac{-250 + 525 - 300}{6} \right] - \left[ \frac{-16 + 84 - 120}{6} \right] = \frac{-25}{6} - \left( \frac{-52}{6} \right) = \frac{-25 + 52}{6} = \frac{27}{6} = \frac{9}{2} \] ### 4. **Final Answer** \[ \text{Area} = \frac{9}{2} \text{ square units} \] --- **Summary:** - **i)** The parabola \( y = (x - 3)^2 \) and the line \( y = x - 1 \) intersect at points \( (2, 1) \) and \( (5, 4) \). The line lies above the parabola between these points. - **ii)** The area of the enclosed region is \( \frac{9}{2} \) square units.